Physics 211: Lecture 5, Pg 1 Physics 211: Lecture 5 Today’s Agenda l More discussion of dynamics  Recap Newton's Laws Free Body Diagram  The Free Body Diagram  The tools we have for making & solving problems:  Ropes & Pulleys (tension)  Hooke’s Law (springs) Original: Mats A.Selen Modified: W.Geist

Physics 211: Lecture 5, Pg 2 Materials l Bring strings and springs. l Two spring powered carts on track: Use equal and unequal masses. Have students predict the outcome (i.e. accelerations and final speeds)

Physics 211: Lecture 5, Pg 3 Review: Newton's Laws Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Fa FF Law 2: For any object, F NET = ma Where F NET =  F action-reactionFF F Law 3: Forces occur in action-reaction pairs, F A,B = - F B,A. Where F A,B is the force acting on object A due to its interaction with object B and vice-versa. m is “mass” of object

Physics 211: Lecture 5, Pg 4 Gravity: Mass vs Weight l What is the force of gravity exerted by the earth on a typical physics student?  Typical student mass m = 55kg  g = 9.81 m/s 2.  F g = mg = (55 kg)x(9.81 m/s 2 )  F g = 540 N = WEIGHT F= -g F E,S = -mg F = F= g F S,E = F g = mg

Physics 211: Lecture 5, Pg 5 Lecture 5, Act 1 Mass vs. Weight l An astronaut on Earth kicks a bowling ball straight ahead and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon in the same manner with the same force. l His foot hurts... (a) more (b) less (c) the same

Physics 211: Lecture 5, Pg 6 Lecture 5, Act 1 Solution l The masses of both the bowling ball and the astronaut remain the same, so his foot will feel the same resistance and hurt the same as before. Ouch.

Physics 211: Lecture 5, Pg 7 Lecture 5, Act 1 Solution l However the weights of the bowling ball and the astronaut are less: l Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. W = mg Moon g Moon < g Earth

Physics 211: Lecture 5, Pg 8 The Free Body Diagram Fa l Newton’s 2nd Law says that for an object F = ma. for an object. l Key phrase here is for an object.  Object has mass and experiences forces Fa l So before we can apply F = ma to any given object we isolate the forces acting on this object:

Physics 211: Lecture 5, Pg 9 The Free Body Diagram... l Consider the following case as an example of this….  What are the forces acting on the plank ?  Other forces act on F, W and E. focus on plank P = plank F = floor W = wall E = earth F F W,P F F P,W F F P,F F F P,E F F F,P F F E,P

Physics 211: Lecture 5, Pg 10 The Free Body Diagram... l Consider the following case  What are the forces acting on the plank ? Isolate the plank from the rest of the world. F F W,P F F P,W F F P,F F F P,E F F F,P F F E,P

Physics 211: Lecture 5, Pg 11 The Free Body Diagram... l The forces acting on the plank should reveal themselves... F F P,W F F P,F F F P,E

Physics 211: Lecture 5, Pg 12 Aside... l In this example the plank is not moving...  It is certainly not accelerating! Fa F  So F NET = ma becomes F NET = 0  This is the basic idea behind statics, which we will discuss in a few weeks. FFF F P,W + F P,F + F P,E = 0 F F P,W F F P,F F F P,E

Physics 211: Lecture 5, Pg 13 Example l Example dynamics problem: x A box of mass m = 2 kg slides on a horizontal frictionless floor. A force F x = 10 N pushes on it in the x direction. What is the acceleration of the box? Fi F = F x i a a = ? m y x

Physics 211: Lecture 5, Pg 14 Example... l Draw a picture showing all of the forces F F F B,F F F F,B F F B,E F F E,B y x

Physics 211: Lecture 5, Pg 15 Example... l Draw a picture showing all of the forces. l Isolate the forces acting on the block. F F F B,F F F F,B F g F B,E = mg F F E,B y x

Physics 211: Lecture 5, Pg 16 Example... l Draw a picture showing all of the forces. l Isolate the forces acting on the block. l Draw a free body diagram. F F F B,F gmggmg y x

Physics 211: Lecture 5, Pg 17 Example... l Draw a picture showing all of the forces. l Isolate the forces acting on the block. l Draw a free body diagram. l Solve Newton’s equations for each component.  F X = ma X  F B,F - mg = ma Y F F F B,F gmggmg y x

Physics 211: Lecture 5, Pg 18 Example... l F X = ma X  So a X = F X / m = (10 N)/(2 kg) = 5 m/s 2. l F B,F - mg = ma Y  But a Y = 0  So F B,F = mg. Normal Force l The vertical component of the force of the floor on the object (F B,F ) is often called the Normal Force (N). l Since a Y = 0, N = mg in this case. FXFX N mg y x

Physics 211: Lecture 5, Pg 19 Example Recap FXFX N = mg mg a X = F X / m y x

Physics 211: Lecture 5, Pg 20 Lecture 5, Act 2 Normal Force (Check your answer: Similar to mass attached to string) l A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block? m (a) N > mg (b) N = mg (c) N < mg a

Physics 211: Lecture 5, Pg 21 Lecture 5, Act 2 Solution m N mg All forces are acting in the y direction, so use: F total = ma N - mg = ma N = ma + mg therefore N > mg a

Physics 211: Lecture 5, Pg 22 Tools: Ropes & Strings l Can be used to pull from a distance. l Tension l Tension (T) at a certain position in a rope is the magnitude of the force acting across a cross-section of the rope at that position.  The force you would feel if you cut the rope and grabbed the ends.  An action-reaction pair. cut T T T Tension doesn’t have a direction. When you hook up a wire to an object the direction is determined by geometry of the hook up.

Physics 211: Lecture 5, Pg 23 Tools: Ropes & Strings... l Consider a horizontal segment of rope having mass m:  Draw a free-body diagram (ignore gravity). x l Using Newton’s 2nd law (in x direction): F NET = T 2 - T 1 = ma So if m = 0 (i.e. the rope is light) then T 1 =  T 2  T is constant anywhere in a rope or string as long as its massless T1T1 T2T2 m ax

Physics 211: Lecture 5, Pg 24 Tools: Ropes & Strings... l An ideal (massless) rope has constant tension along the rope. l If a rope has mass, the tension can vary along the rope  For example, a heavy rope hanging from the ceiling... l We will deal mostly with ideal massless ropes. T = T g T = 0 TT 2 skateboards

Physics 211: Lecture 5, Pg 25 Tools: Ropes & Strings... l What is force acting on box by the rope in the picture below? (always assume rope is massless unless told different) mg T m Since a y = 0 (box not moving), T = mg

Physics 211: Lecture 5, Pg 26 Lecture 5, Act 3 Force and acceleration l A fish is being yanked upward out of the water using a fishing line that breaks when the tension reaches 180 N. The string snaps when the acceleration of the fish is observed to be is 12.2 m/s 2. What is the mass of the fish? m = ? a = 12.2 m/s 2 snap ! (a) 14.8 kg (b) 18.4 kg (c) 8.2 kg For what object do you draw the FBD?

Physics 211: Lecture 5, Pg 27 Lecture 5, Act 3 Solution: l Draw a Free Body Diagram!! T mg m = ? a = 12.2 m/s 2 l Use Newton’s 2nd law in the upward direction: F TOT = ma T - mg = ma T = ma + mg = m(g+a)

Physics 211: Lecture 5, Pg 28 Tools: Pegs & Pulleys l Used to change the direction of forces  An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: FF1FF1 ideal peg or pulley FF2FF2 FF | F 1 | = | F 2 |

Physics 211: Lecture 5, Pg 29 Tools: Pegs & Pulleys l Used to change the direction of forces  An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: mg T m T = mg F W,S = mg

Physics 211: Lecture 5, Pg 30 Springs l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.  F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position F X = 0 x

Physics 211: Lecture 5, Pg 31 Springs... l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.  F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position F X = -kx > 0 x x  0

Physics 211: Lecture 5, Pg 32 Springs... l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.  F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. F X = - kx < 0 x x > 0 relaxed position Horizontal springs

Physics 211: Lecture 5, Pg 33 Scales: l Springs can be calibrated to tell us the applied force.  We can calibrate scales to read Newtons, or...  Fishing scales usually read weight in kg or lbs lb = 4.45 N Spring/string

Physics 211: Lecture 5, Pg 34 mmm (a) (b) (c) (a) 0 lbs. (b) 4 lbs. (c) 8 lbs. (1) (2) ? Lecture 5, Act 4 Force and acceleration l A block weighing 4 lbs is hung from a rope attached to a scale. The scale is then attached to a wall and reads 4 lbs. What will the scale read when it is instead attached to another block weighing 4 lbs? Scale on a skate

Physics 211: Lecture 5, Pg 35 Lecture 5, Act 4 Solution: l Draw a Free Body Diagram of one of the blocks!! l Use Newton’s 2nd Law in the y direction: F TOT = 0 T - mg = 0 T = mg = 4 lbs. mg T m T = mg a = 0 since the blocks are stationary

Physics 211: Lecture 5, Pg 36 Lecture 5, Act 4 Solution: l The scale reads the tension in the rope, which is T = 4 lbs in both cases! mmm T T T T T T T

Physics 211: Lecture 5, Pg 37 Recap of today’s lecture.. l More discussion of dynamics (Read Chapter 4 in text) èRecap Free Body Diagram èThe Free Body Diagram èThe tools we have for making & solving problems: »Ropes & Pulleys (tension) »Hooke’s Law (springs).