Dynamics Problems.

Dynamics Problems

Warm Up The system below is in equilibrium. If the scale is calibrated in N, what does it read? Since the tension is distributed over the entire (mass-less) rope, the scale will read (9.8N/kg)(5kg) = 49N 5kg 5kg

Warm Up Determine the Normal force in each of the following: F W F F m

Warm Up Write F=ma for each scenario: m F a F W F m

Example Diagram Free body Diagram
A pumpkin of unknown mass is suspended by a cord attached to the ceiling and pushed away from vertical. When a 24.0 N force is applied to the pumpkin at an angle of to horizontal, the pumpkin will remain in equilibrium when the cord makes an angle of with the vertical. (A) What is the tension in the cord when the pumpkin is in equilibrium ? (B) What is the mass of the pumpkin ? Diagram Free body Diagram

(A) What is the tension in the cord when the pumpkin is in equilibrium ?
Since the pumpkin is in static equilibrium, we need only look at the horizontal components

(B) What is the mass of the pumpkin ?
Since the pumpkin is in static equilibrium, we need only look at the vertical components

Example Diagram Free body Diagram
The tension in the horizontal rope is 30N A) Determine the weight of the object Diagram Free body Diagram 30N 400 500 500

A) Determine the weight of the object
500 The weight is in static equilibrium, so the appropriate net component forces must be zero. Horizontal Component Vertical Component The weight of the mass is 36N

Weight on a Wire Diagram Free body Diagram
A rope extends between two poles. A 80N weight hangs from it as per the diagram. A) Determine the tension in both parts of the rope. Diagram Free body Diagram 100 150 80N T1 T2

A) Determine the tension in both parts of the rope.
The weight is in static equilibrium, so the appropriate net component forces must be zero. Horizontal Component Vertical Component

Example Free body Diagram Block Ring
The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a ring, what is the coefficient of static friction between the block and the tabletop?. Free body Diagram Block Ring 35N 7.0N

The system in the diagram is just on the verge of slipping. If a 7
The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a ring, what is the coefficient of static friction between the block and the tabletop?. The weight is just in static equilibrium, so the appropriate net component forces must be zero. Ring Block From Block: From ring: Combining:

Atwood’s Machine Example:
Masses m1 = 10 kg and m2 = 20kg are attached to an ideal massless string and hung as shown around an ideal massless pulley. What are the tensions in the string T1 and T2 ? Find the accelerations, a1 and a2, of the masses. Fixed Pulley T1 T2 m1 a1 m2 a2

Try These

Draw free body diagrams for each object
Applying Newton’s Second Law: T1 - m1g = m1a (a) T2 - m2g = -m2a2 => -T2 + m2g = +m2a2 (b) But T1 = T2 = T since pulley is ideal and a1 = -a2 =a since the masses are connected by the string m2g m1g Free Body Diagrams T1 T2 a1 a2

Solve for Acceleration
-m1g + T = m1 a (a) -T + m2g = m2 a (b) Two equations and two unknowns we can solve for both unknowns (T and a). Add (b) + (a): g(m2 – m1 ) = a(m1+ m2 ) m2g m1g Free Body Diagrams T1 T2 a1 a2

Solve for T -m1g + T = m1 a (a) -T + m2g = m2 a (b)
Plug a into (b) and Solve for T m2g m1g Free Body Diagrams T1 T2 a1 a2

Atwood Machine Review So we find: m1 m2 a T

Example A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box?

Pulling a Box (Part 1) A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box? Free body Diagram +y +x a

Pulling a Box (Part 1) A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box? +y +x a Vertical Forces Horizontal Forces Solving

Example Two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes?

Pulling a Box (Part 2) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? Free body Diagram +y +x a

Pulling a Box (Part 2) + Forces
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? a +y +x Forces 4.00 kg Box 6.00 kg Box Adding to eliminate T and find a +

Pulling a Box (Part 2) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? a +y +x Solve for Acceleration Now for Tension We could have used the other tension formula from Box 2 and obtained the same answer

Example A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is connected by a 1kg rope and they are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? 1.00kg

Pulling a Box (Part 3) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? Free body Diagram 1.00 kg Because the rope has mass, the two ends will experience different tensions +y +x a

Pulling a Box (Part 3) Forces
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? a +y +x Forces 4.00 kg Box 6.00 kg Box Using F=ma for the system to find a

Pulling a Box (Part 3) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? a +y +x Solve for Acceleration Now for T1 Now for T2

Example A worker drags a 38.0 kg box along the floor by pulling on a rope attached to the box. The coefficient of friction between the floor and the box is us=0.450 and uk=0.410. a) What are the force of friction and acceleration when the worker applies a horizontal force of 150N? b) What are the force of friction and acceleration when the worker applies a horizontal force of 190N? 38kg

Solution (Free Body Diagram)
a) What are the force of friction and acceleration of the worker applies a horizontal force of 150N? The Normal force up Friction to the left The applied force of tension to the right 38kg The force of gravity down +y +x

Solution (Vector Components)
What are the force of friction and acceleration of the worker applies a horizontal force of 150N? 38kg +y +x To determine if the box will move, we must find the maximum static friction and compare it to the applied force. Since the applied force by the worker is only 150N, the box will not move

Solution (Vector Components)
b) What are the force of friction and acceleration of the worker applies a horizontal force of 190N? 38kg +y +x Since the applied force is greater than 168N from part a), we will have an acceleration in the x direction. So we will apply Newton’s 2nd Law in the horizontal direction. FK=(0.410)(372.4N)=153N The acceleration of the box is m/s2 [E]

Example A train of three masses is pulled along a frictionless surface. Calculate the tensions in the ropes. T1 T2 8 kg 5 kg 13 kg 30 N We can find the acceleration of the train by treating the three masses as one unit. T2 T1 Tension in rope T2 or T1 F Tension in rope T1

Example A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40? 300

A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40? The Normal force up Friction to the left The applied force of tension at 300 300 Tension broken down into components The force of gravity down

Solution (Force Components)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40? 300 +y +x Vertical Components Horizontal Components Since we have a constant velocity, acceleration is 0 We will need FN, so solve for FN

Solution (Force Components)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40? 300 +y +x Solve for FT

Example A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s.

A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. +x +y Choose axis orientation to match the direction of motion and the normal to the surface Normal is perpendicular to the surface Object Decompose gravity into axis components Force of friction opposes direction of motion Force of gravity is straight down

Remember to solve for FN because we will need it later
Solution (Force Vectors) A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. +x +y y direction x direction Remember to solve for FN because we will need it later

Example 9: Solution (Force Vectors)
A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. +x +y Acceleration Speed

Example Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley.

Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. +y +x a Box B Normal Force Friction from A Friction from Table Object B Force of Gravity from A and B Tension Applied Force

Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. +y +x a Normal Force Box A Friction Object Tension Force of Gravity from A only +y +x a

Solution (Force Vectors)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. A B a +y +x x-direction for Box A x-direction for Box B This was determined using Box A

Solution (Force Vectors)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. A B a +y +x

Stacked Blocks with Friction
Two blocks are stacked on top of each other and both are at rest. The coefficient of static friction between the blocks is 0.25 and the coefficient of kinetic friction is Block A rests upon a frictionless surface. A horizontal force of 50 N is then applied to the bottom box. At what rate does each box accelerate? This type of problem is trickier than it first appears. The problem is with the friction between the two boxes. Is that friction enough to hold them together so they will slide as one, or will the 5kg box slide over the 10kg box?

Stacked Blocks with Friction -Solution
Two blocks are stacked on top of each other and both are at rest. The coefficient of static friction between the blocks is 0.25 and the coefficient of kinetic friction is Block A rests upon a frictionless surface. A horizontal force of 50 N is then applied to the bottom box. At what rate does each box accelerate? Let’s first determine at what acceleration rate, the 5 kg box can stay attached to the 10 kg box due to static friction. Note: since the Force is to the right, so then is the static force on the top block from the bottom block. This 2.45 m/s2 (maximum) acceleration corresponds to a Force of (15kg)(2.45m/s2) or 36.75N on the system. Any force less than or equal to 36.75N and we can treat the system as one. Therefore any force that produces an acceleration larger than 2.45 m/s2, will cause the 5kg block to slide, and we will have to use the coefficient of kinetic friction to calculate its acceleration.

Two blocks are stacked on top of each other and both are at rest. The coefficient of static friction between the blocks is 0.25 and the coefficient of kinetic friction is Block A rests upon a frictionless surface. A horizontal force of 50 N is then applied to the bottom box. At what rate does each box accelerate? So, what is the acceleration that the system experiences with the given conditions? Let’s treat them as one big 15 kg block and find out. Hey, this is greater that the maximum acceleration of 2.45 m/s2 . So the force of the Static friction is not strong enough to hold them together, and they two blocks will slide (being restricted now by Kinetic Friction). We now can use this information to determine the acceleration of each of the blocks.

Two blocks are stacked on top of each other and both are at rest. The coefficient of static friction between the blocks is 0.25 and the coefficient of kinetic friction is Block A rests upon a frictionless surface. A horizontal force of 50 N is then applied to the bottom box. At what rate does each box accelerate? Acceleration of the 5 kg Block. Therefore the 5 kg block will accelerate to the right at about 1.8 m/s2.

Two blocks are stacked on top of each other and both are at rest. The coefficient of static friction between the blocks is 0.25 and the coefficient of kinetic friction is Block A rests upon a frictionless surface. A horizontal force of 50 N is then applied to the bottom box. At what rate does each box accelerate? We not that the force of kinetic friction due to the top box is to the left on the bottom box. Acceleration of the 10 kg Block. FN Ff Fapplied=50N Fg=m5g Fg=m10g Therefore the 5 kg block will accelerate to the right at about 4.1 m/s2.

Multiple blocks Determine the acceleration a1, a2, and a3 of the three blocks if a horizontal force of 10N is applied on: The 2 kg block The 3 kg block The 7 kg block

Multiple blocks Determine the acceleration a1, a2, and a3 of the three blocks if a horizontal force of 10N is applied on: The 2 kg block a1: Ff12 2kg Fapplied=10N Note that the Applied Force is stronger than the Maximum Friction Force, so it slides. Therefore the acceleration on the 2kg block is about 3 m/s2

Multiple blocks Determine the acceleration a1, a2, and a3 of the three blocks if a horizontal force of 10N is applied on: The 2 kg block Since the 3kg and the 7kg blocked at fixed and there is no friction beneath the 7kg Block, we can treat them as a 10kg block and redo the question. FN a2: Ff23 3kg Ff12=3.92N Fg=2g Fg=3g Note that the Applied Friction Force (3.9N) is weaker than the Maximum Retarding Friction Force (14.7N), so the two Blocks remain attached and there is no relative motion between the 3kg and the 7kg Blocks). Therefore both a2 and a3 have an acceleration of about 0.4 m/s2

Multiple blocks Determine the acceleration a1, a2, and a3 of the three blocks if a horizontal force of 10N is applied on: b) The 3 kg block FN Since the 2kg, 3kg and the 7kg blocked at fixed and there is no friction beneath the 7kg Block, we can treat them as a 12kg block and redo the question. 3.92N=Ff12 a2: 3kg 14.7N=Ff23 FApplied=10N Fg=2g Fg=3g Note that the combined Maximum Applied Friction Force (18.6N) is stronger than the Applied Force (10N), so the three Blocks remain attached and there is no relative motion between the them. Therefore a1, a2 and a3 have an acceleration of about m/s2

Multiple blocks Determine the acceleration a1, a2, and a3 of the three blocks if a horizontal force of 10N is applied on: c) The 7 kg block Ff23 7kg a3: Fapplied=10N Note that the Applied Force is weaker than the Maximum Friction Force, so there is no relative motion, so they stick together. Then as is b) the 2kg block also sticks to the 3kg Block, that is they all remain attached. Therefore a1, a2 and a3 have an acceleration of about m/s2

Example How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 1.5 kg 2.0 1.0

Example (free body diagram)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 3 Blocks taken as a Single Unit Normal Force 1.5 kg 2.0 1.0 1.5 kg 2.0 1.0 Friction Object Applied +y +x Force of Gravity a

Example (force vectors)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 1.5 kg 2.0 1.0 +y +x a

Example (free body diagram)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 2.0 kg Block and 1.0 kg taken as a Single Unit Since we are considering 2.0kg and 1.0kg block as a unit, then the Force is the push of 1.5 kg block on the combined block Normal Force Friction Object 2.0 kg Applied 1.5 kg 1.0 kg Force of Gravity +y +x a

Example (force vectors)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 2.0 kg 1.0 +y +x a

Example Step 1: Free body Diagram FT FT m1 + FG=9 kg +
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object. Find the acceleration of the two objects Find the tension in the string. Step 1: Free body Diagram FG=9 kg FT FT m1 + + The easiest way to choose the signs for the forces is to logical choose what you believe will be correct direction and follow that direction from one object to the other. If your final answer is negative, it just means your initial direction choice was wrong.

Example (Solution) FT FT m1 + FG=9 kg +
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object. Find the acceleration of the two objects FG=9 kg FT FT m1 + + Treat this object as one object with forces pulling to left and pulling to right.

Example (Solution) FT FT m1 +
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object. Find the tension in the string. FG=9 kg FT FT m1 + We need only substitute the acceleration value into either the horizontal or vertical equation. Use the Horizontal since only one force is being applied + Horizontal Vertical

Example Blocks A, B, and C are placed as in the figure and connected by ropes of negligible mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block C descends with constant velocity. Determine the tension in the rope connecting Block A and B. What is the weight of Block C? If the rope connecting A and B were cut, what would be the acceleration of C?

Example Normal Block A Block B Normal Tension from C Object Tension
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity. Determine the tension in the rope connecting Block A and B. What is the weight of Block C? If the rope connecting A and B were cut, what would be the acceleration of C? Normal Block A Block B Normal Tension from C Object +x +y Tension Object Friction Tension from A Gravity Gravity Friction

Example (Force Vectors)
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity. Determine the tension in the rope connecting Block A and B. What is the weight of Block C? If the rope connecting A and B were cut, what would be the acceleration of C? Block A Block C Block B Let’s just isolate Block A

Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity. Determine the tension in the rope connecting Block A and B. What is the weight of Block C? If the rope connecting A and B were cut, what would be the acceleration of C? Let’s treat them all as one object with forces to the left and right.

Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity. Determine the tension in the rope connecting Block A and B. What is the weight of Block C? If the rope connecting A and B were cut, what would be the acceleration of C? Block A Block C Block B

Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity. Determine the tension in the rope connecting Block A and B. What is the weight of Block C? If the rope connecting A and B were cut, what would be the acceleration of C? When the rope is cut:

Example A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration.

Example (Free Body Diagram)
A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration. We will look at this from outside the truck (ie the ground) because we would prefer an inertial frame of reference. Tension broken into components Tension Object Gravity

A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration. Vertical Forces Horizontal Forces

Example Calculate the acceleration of a box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal. 5.0 kg 25N

Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal. Since the gravity force down (5x9.8) is greater than force up (25sin(30), the box slides down, so friction is up. Applied Friction 5.0 kg 25N Object Magnetic Normal Gravity

Example (Vector Forces)
Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal. 5.0 kg 25N +y +x Vertical Horizontal

Example (Insert Numbers)
Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal. 5.0 kg 25N +y +x

Example Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 604N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15.

Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 604N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15. Treat objects 300 thru 121 as a single mass of 180*0.5kg For Masses 121 to 300 Normal Big Mass Tension from above Gravity Friction

Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 604N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15. For Combined Masses 121 to 300 a +y +x y-axis x-axis

Example (Insert Values)
Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 604N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15. For Combined Masses 1 to 120 a +y +x

Example A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. 300 a

A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. 300 FN a Ff 300 +y +x Fa Fg

A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. +y +x Fg FN Ff Fa 300 a y-axis Insert Values

A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. Fg FN Ff Fa 300 +y +x a Friction

A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. Fg FN Ff Fa 300 +y +x a Acceleration

Example Calculate the unknowns for each accelerated block. b) a) a c)
18 kg a) F 6 kg b) a m c)

Example (Solution) Calculate the unknowns for each accelerated block.
18 kg a) F

6 kg b) a

Example Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μs

The only horizontal force is the Normal Force .Therefore F=N=ma
Example Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μs If I do not fall, then the friction force, Ff, must balance my weight mg, that is Ff = mg The only horizontal force is the Normal Force .Therefore F=N=ma Putting this together we obtain:

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is uk=0.15, and both masses are 3.0 kg. a) If m2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of uk, that will keep the system from accelerating?

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is uk=0.15, and both masses are 3.0 kg. a) If m2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of uk, that will keep the system from accelerating? m2 m1 FN FT FT Ff Fg=m2g Fg=m1g

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is uk=0.15, and both masses are 3.0 kg. a) If m2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of uk, that will keep the system from accelerating? m2 Fg=m2g FT m1 FN Fg=m1g Because the two masses are connected, we can treat them as one unit and just apply the forces that move it one way or another.

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is uk=0.15, and both masses are 3.0 kg. a) If m2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of uk, that will keep the system from accelerating? m2 Fg=m2g FT m1 FN Fg=m1g

Attached bodies on two inclined planes
+ m1 y x m2 T1 N m1g 1 m2g T2 2 Step 1 Free Body Diagram Pick a direction in which you think the blocks will move and make that direction positive

We want to eliminate T, so let’s add
y x m2 T1 N m1g 1 m2g T2 2 Block 1 Block 2 We want to eliminate T, so let’s add + Since the acceleration is negative, our original choice for positive direction was wrong, so mass 1 goes up, and mass 2 goes down both at 2.2 m/s2.

Two-body dynamics Case (1) Case (2)
In which case does block m experience a larger acceleration? In (1) there is a 10 kg mass hanging from a rope. In (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless. m 10kg a a m F = 98.1 N Case (1) Case (2) (a) Case (1) (b) Case (2) (c) same

Solution For case (1) draw FBD and write FNET = ma for each block: (a) T = ma (a) mWg -T = mWa (b) m 10kg a Add (a) and (b): mWg = (m + mW)a mW=10kg (b) Note:

Solution T = 98.1 N = ma Case (1) Case (2) For case (2) m m a a 10kg
F = 98.1 N Case (2) The answer is (b) Case (2) In this case the block experiences a larger acceleratioin

Bucket Problem A Lady in a bucket (don’t ask me why she is in a bucket), with combined mass of 70 kg),pulls herself upward by pulling down on a rope (which is attached through a pulley to the bucket). How hard must she pull down on the rope to raise herself at a constant rate (remember remaining still is at a constant rate of 0 m/s) If she increases this force by 25%, how fast will she accelerate up?

Bucket Problem A Lady in a bucket (don’t ask me why she is in a bucket), with combined mass of 70 kg),pulls herself upward by pulling down on a rope (which is attached through a pulley to the bucket). How hard must she pull down on the rope to raise herself at a constant rate (remember remaining still is at a constant rate of 0 m/s) In these types of questions, it is best to treat Lady and the Bucket as one object. As the Lady pulls down on the rope, the rope pulls up on her hands. Just think of her holding the bucket still.

Bucket Problem A Lady in a bucket (don’t ask me why she is in a bucket), with combined mass of 70 kg), pulls herself upward by pulling down on a rope (which is attached through a pulley to the bucket). If she increases this force by 25%, how fast will she accelerate up?

Understanding A person standing on a horizontal floor feels two forces: the downward pull of gravity and the upward supporting force from the floor. These two forces Have equal magnitudes and form an action/reaction pair Have equal magnitudes but do not form an action/reaction pair Have unequal magnitudes and form an action/reaction pair Have unequal magnitudes and do not form an action/reaction pair None of the above Because the person is not accelerating, the net force they feel is zero. Therefore the magnitudes must be the same (opposite directions. These are not action/reaction forces because they act of the same object (the person). Action/Reaction pairs always act on different objects.

Dynamics Problems.

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Dynamics Problems.

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