Two-Body Pulley Problems. Consider the following system: M A = 3.0 kg M B = 1.5 kg Pulley a) frictionless b) μ K = 0.200 table A B.

Slides:



Advertisements
Similar presentations
FORCE A force is any influence that can change the velocity of a body. Forces can act either through the physical contact of two objects (contact forces:
Advertisements

No Strings Attached: Normal Forces, Force Vectors, Strings, Springs and Pulleys Chapter Important Vocabulary: Normal Force Contact Force Tension.
Newton’s Laws of Motion and Free Body Analysis
Chapter 7. Newton’s Third Law
Chapter 4 The Laws of Motion.
Forces and Newton’s Laws of Motion Chapter 4. All objects naturally tend to continue moving in the same direction at the same speed. All objects resist.
Newton’s Laws and Gravity Mass, Weight, Force, Friction. Application of Newton’s Laws o Introduce Newton’s three laws of motion  At the heart of classical.
Follow the same procedure as other force problems, but keep in mind: 1) Draw a free body diagram for EACH object or for each junction in a rope.
Applying Forces (Free body diagrams).
Problem Block A of mass 12 kg and block B of mass 6 kg are connected by a cable that passes over pulley C which can rotate freely. Knowing that the.
04-1 Physics I Class 04 Newton’s Second Law for More Complex Cases.
Lecture 4 Monday: 26 January 2004 Newton’s Laws of Motion.
Physics 101: Lecture 9, Pg 1 Physics 101: Application of Newton's Laws l Review of the different types of forces discussed in Chapter 4: Gravitational,
Physics 101: Lecture 9, Pg 1 Physics 101: Application of Newton's Laws l Review of the different types of forces discussed in Chapter 4: Gravitational,
Newton’s Laws of Motion Problems MC Questions
Copyright © 2010 Pearson Education South Asia Pte Ltd
NEWTON'S LAWS OF MOTION There are three of them.
Newton’s Laws Problems
Forces.
AP Physics B Summer Course 年AP物理B暑假班
Chapter 6 Force and Motion.
Forces Contact Forces - those resulting from physical contact between objects –Normal Force –Friction –Tension (spring/rope) –Compression Action at a Distance.
Aim: More Law of Inertia
Newton’s Laws of Motion
Solving Problems with Newton’s Laws
AP Physics I.B Newton’s Laws of Motion. B.1 An interaction between two bodies resulting in a push or a pull is a force. Forces are of two types: contact.
Equilibrium of a particle
What is the normal force for a 500 kg object resting on a horizontal surface if a massless rope with a tension of 150 N is acting at a 45 o angle to the.
Newton’s Laws - continued Friction, Inclined Planes, N.T.L., Law of Gravitation.
SECOND LAW OF MOTION If there is a net force acting on an object, the object will have an acceleration and the object’s velocity will change. Newton's.
Chapter 4 Dynamics: Newton’s Laws of Motion
 Isaac Newton  Smart Guy  Liked Apples  Invented Calculus  Came up with 3 laws of motion  Named stuff after himself.
1 Some application & Forces of Friction. 2 Example: When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass,
Physics 1D03 - Lecture 71 Newton’s Laws (II) Free-body diagrams Normal Force Friction.
Frictional Forces  Two types: - static – applies to stationary objects - kinetic – applies to sliding (moving) objects  Like F N, the Frictional Force.
Newton 2nd Law problems - Atwood Machines -Incline Planes -Tension Problems -Other Object Connected problems.
Torque Calculations with Pulleys
Aim: More Atwood Machines Answer Key HW 6 Do Now: Draw a free-body diagram for the following frictionless inclined plane: m2m2 m1m1 M θ Mg m2m2 m1m1 M.
Unit 2 1D Vectors & Newton’s Laws of Motion. A. Vectors and Scalars.
Forces on Inclined Planes Unit 3, Presentation 3.
Forces and Angles Physics 11.
Lecture 9 Serway and Jewett : 5.7, 5.8
Ch 4. Forces & Newton’s Laws of Motion
Multiple Object Systems 1. Analyze the system as one object. 2. Analyze each object individually. 3. Create multiple equations to solve for multiple unknowns.
Physics – Chapter 4 Pulleys
5a. The plane and pulley are frictionless a) If A has a mass of 23.1 kg, and B has a mass of 5.63 kg, what is the tension in the string, and the acceleration.
Newton’s Laws.
Free-Body Diagrams PHYSICS SEMESTER ONE NANSLO Physics Core Units and Laboratory Experiments by the North American Network of Science Labs Online,North.
Applications of Newton’s Laws Physics 11. Numerous Applications Apparent weight Free fall Inclined planes Atwood’s machines Universal Law of Gravitation.
University Physics: Mechanics
Newton’s Three Laws of Motion. Newton’s 2 nd Law.
Net Force Physics 11. Comprehension Check 1. An object has a weight of 12.2N; what is its mass? 2. If you were to take the weight of a 1.0kg object on.
“ Friendship is like peeing on yourself; everyone can see it, but only you get the warm feeling it brings.” Funnyquotes.com Course web page
Inclined Plane Problems. Axes for Inclined Planes X axis is parallel to the inclined plane Y axis is perpendicular to the inclined plane Friction force.
Force Problems. A car is traveling at constant velocity with a frictional force of 2000 N acting opposite the motion of the car. The force acting on the.
What is a force? An interaction between TWO objects. For example, pushes and pulls are forces. We must be careful to think about a force as acting on one.
Unit is the NEWTON(N) Is by definition a push or a pull Can exist during physical contact(Tension, Friction, Applied Force) Can exist with NO physical.
Dynamics and Relative Velocity Lecture 07 l Problem Solving: Dynamics Examples l Relative Velocity.
Acceleration of a Single Body Sliding Down an Incline
Lecture 2: More Forces.
Atwood Machines and Multiple Body Systems
Free Body Diagrams.
Applications of Newton’s Laws Tension and Pulleys
Compound Body Problems
4-6 Weight – the Force of Gravity; and the Normal Force
How does an inclined plane make work easier How does an inclined plane make work easier? How does it change the force that is applied to the inclined.
Applying Forces AP Physics 1.
NEWTON'S LAWS OF MOTION There are three of them.
Applying Forces AP Physics C.
NEWTON'S LAWS OF MOTION There are three of them.
Presentation transcript:

Two-Body Pulley Problems

Consider the following system: M A = 3.0 kg M B = 1.5 kg Pulley a) frictionless b) μ K = table A B

Two-Body Pulley Problems Consider the following system: What is a pulley? M A = 3.0 kg M B = 1.5 kg Pulley a) frictionless b) μ K = table A B

Two-Body Pulley Problems Consider the following system: M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B

Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B

Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = Step One : ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B

Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = Step One : determine the direction of the acceleration of each body and label on the diagram. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B

Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = Step One : determine the direction of the acceleration of each body and label on the diagram. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB

Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = Step One : determine the direction of the acceleration of each body and label on the diagram. Step Two: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB

Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = Step One : determine the direction of the acceleration of each body and label on the diagram. Step Two: compare the magnitude of the acceleration of each block M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB

Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = Step One : determine the direction of the acceleration of each body and label on the diagram. Step Two: compare the magnitude of the acceleration of each block If block B descends by 1 cm, it is clear that block A will move to the right by 1 cm. (conservation of rope) Therefore | a A | = | a B | = a Green a will represent the positive size of the acceleration. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB

Two-Body Pulley Problems Step 3: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB

Two-Body Pulley Problems Step 3: Draw an FBD of block A M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Step 3: Draw an FBD of block A M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = ? FNFN A

Two-Body Pulley Problems Step 3: Draw an FBD of block A M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces What is a y = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces a y = 0 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces a y = 0 According to Newton's first law, what is F net y = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement F N - 30 = 0 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement F N - 30 = 0 F N = 30 N [up ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement F N - 30 = 0 F N = 30 N [up ] Label on FBD M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement F N - 30 = 0 F N = 30 N [up ] Label on FBD M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement ???? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T = 3 a Scalar Statement ???? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T = 3 a Scalar Statement T = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T = 3 a Scalar Statement T = 3 aequation #1 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems Step 7b: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B aBaB T FgFg A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B aBaB T F g = ? A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B aBaB T F g = 15 N A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B aBaB T F g = 15 N A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B aBaB T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B aBaB T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Vector statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Vector statement T + F g = 1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Vector statement T + F g = 1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement ??? Vector statement T + F g = 1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement T - 15 = -1.5 a Vector statement T + F g = 1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement T - 15 = -1.5 a Vector statement T + F g = 1.5 aT = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement T - 15 = -1.5 a Vector statement T + F g = 1.5 aT = a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement T - 15 = -1.5 a Vector statement T + F g = 1.5 aequation #2 T = a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 a = 3.3 m/s 2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 a = 3.3 m/s 2 a A = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] How can we find the magnitude of the tension force in the string? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] Back -sub into #1 T = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] Back -sub into #1 T = 3 a tension in string T = 3 ( 3.3) = 9.9 N M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = a equation #2 Step 11: Solve by sub #1 into #2 3 a = a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] Back -sub into #1 T = 3 a tension in string T = 3 ( 3.3) = 9.9 N M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB Step 12 Conclusion: For part a) where there is no friction between block A and the table, a A = 3.3 m/s 2 [ right ] and a B = 3.3 m/s 2 [ down ] and the tension in the string is 9.9 N A

Two-Body Pulley Problems Now let's look at part b): Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB

Two-Body Pulley Problems Now let's look at part b): Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = Note | a A | = | a B | = a as before. Which free-body diagram will need to be modified in part b) ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB

Two-Body Pulley Problems Now let's look at part b): Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = Note | a A | = | a B | = a as before. The FBD for block A will need to be modified because kinetic friction now acts. The equation of motion developed for block A will also change. The FBD for block B will not change and the equation of motion developed for block B will still be valid. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB

Two-Body Pulley Problems What addition must be made to our block A FBD ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

Two-Body Pulley Problems What addition must be made to our block A FBD ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y fkfk A

Two-Body Pulley Problems What addition must be made to our block A FBD ? Does F N change as a result of adding f k to our FBD? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y fkfk A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y fkfk A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = ? A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 aT = ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 aT = 3 a + 6 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 aT = 3 a + 6 new equation #1 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a 4.5 a = 9 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a 4.5 a = 9 a = 2.0 m/s 2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a 4.5 a = 9 a = 2.0 m/s 2 a A = ____ m/s 2 [ ____ ] a B = ____ m/s 2 [ ____ ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = ____ m/s 2 [ ____ ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] Back-sub in equation #1 to get T M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] Back-sub in equation #1 to get T T = 3 ( 2.0) + 6 = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] Back-sub in equation #1 to get T T = 3 ( 2.0) + 6 = 12 N M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB A

Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] Back-sub in equation #1 to get T T = 3 ( 2.0) + 6 = 12 N M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = table B aAaA aBaB Conclusion: For part b) where there is friction between block A and the table, a A = 2.0 m/s 2 [ right ] and A B = 2.0 m/s 2 [ down ] and the tension in the string is 12 N A

A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? M A = 3.0 kg M B = 1.5 kg Pulley table A B

A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N 15 ≤ μ s 30 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N 15 ≤ μ s 30 μ s 30 ≥ 15 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N 15 ≤ μ s 30 μ s 30 ≥ 15 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N μ s ≥ 15/30 15 ≤ μ s 30 μ s 30 ≥ 15 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N μ s ≥ 15/30 15 ≤ μ s 30μ s ≥ 0.50 μ s 30 ≥ 15 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N μ s ≥ 15/30 15 ≤ μ s 30μ s ≥ 0.50 μ s 30 ≥ 15The minimum coefficient of static friction is 0.50 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N