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Copyright © 2012 Pearson Education Inc. Application of Newton’s laws: free body diagram Physics 7C lecture 03 Thursday October 3, 8:00 AM – 9:20 AM Engineering Hall 1200

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Copyright © 2012 Pearson Education Inc. Newton’s laws Newton’s First Law of Motion: when forces are balanced the object will keep its motion/velocity. Newton’s Second Law of Motion: F = m a, force causes changes to the motion/velocity. Newton’s Third Law of Motion: action-reaction pairs Yet: The laws are simple to state but intricate in their application. Let’s do some iClicker questions first.

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© 2012 Pearson Education, Inc. Q4.1 v Motor Cable Elevator An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable is A. greater than the downward force of gravity. B. equal to the force of gravity. C. less than the force of gravity. D. any of the above, depending on the speed of the elevator.

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© 2012 Pearson Education, Inc. A4.1 v Motor Cable Elevator An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable is A. greater than the downward force of gravity. B. equal to the force of gravity. C. less than the force of gravity. D. any of the above, depending on the speed of the elevator.

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© 2012 Pearson Education, Inc. Q4.2 v Motor Cable Elevator An elevator is being lowered at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable is A. greater than the downward force of gravity. B. equal to the force of gravity. C. less than the force of gravity. D. any of the above, depending on the speed of the elevator.

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© 2012 Pearson Education, Inc. A4.2 v Motor Cable Elevator An elevator is being lowered at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable is A. greater than the downward force of gravity. B. equal to the force of gravity. C. less than the force of gravity. D. any of the above, depending on the speed of the elevator.

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© 2012 Pearson Education, Inc. Q4.5 A. crate A exerts more force on crate B than B exerts on A B. crate A exerts less force on crate B than B exerts on A C. crate A exerts as much force on crate B as B exerts on A D. Answer depends on the details of the friction force If the two crates are accelerating to the right, A B F A lightweight crate (A) and a heavy crate (B) are side-by-side on a horizontal floor. You apply a horizontal force F to crate A. There is friction between the crates and the floor.

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© 2012 Pearson Education, Inc. A4.5 A. crate A exerts more force on crate B than B exerts on A B. crate A exerts less force on crate B than B exerts on A C. crate A exerts as much force on crate B as B exerts on A D. Answer depends on the details of the friction force If the two crates are accelerating to the right, A B F A lightweight crate (A) and a heavy crate (B) are side-by-side on a horizontal floor. You apply a horizontal force F to crate A. There is friction between the crates and the floor.

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© 2012 Pearson Education, Inc. Q4.6 v Motor Cable Elevator An elevator is being lowered at constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable has the same magnitude as the force of gravity on the elevator, but points in the opposite direction. Why? A. Newton ’ s first law B. Newton ’ s second law C. Newton ’ s third law

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© 2012 Pearson Education, Inc. A4.6 v Motor Cable Elevator An elevator is being lowered at constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable has the same magnitude as the force of gravity on the elevator, but points in the opposite direction. Why? A. Newton ’ s first law B. Newton ’ s second law C. Newton ’ s third law

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© 2012 Pearson Education, Inc. Q4.3 The graph to the right shows the velocity of an object as a function of time. Which of the graphs below best shows the net force versus time for this object? t vxvx 0 A.B.C.D.E. t FxFx 0 t FxFx 0 t FxFx 0 t FxFx 0 t FxFx 0

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© 2012 Pearson Education, Inc. A4.3 The graph to the right shows the velocity of an object as a function of time. Which of the graphs below best shows the net force versus time for this object? t vxvx 0 A.B.C.D.E. t FxFx 0 t FxFx 0 t FxFx 0 t FxFx 0 t FxFx 0

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Copyright © 2012 Pearson Education Inc. Free-body diagrams A free-body diagram is a sketch showing all the forces acting on an object. Why do we need it? What is the tension force in the rope?

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Copyright © 2012 Pearson Education Inc. Free-body diagrams We need to isolate the objects and study the forces on each of them. T1 W1 T1 W2 T2 W3 Note: W = m g action-reaction pairs

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Copyright © 2012 Pearson Education Inc. Free-body diagrams We can apply the 2 nd law to each of the following, or to the group. T1 W1 T1 W2 T2 W3 The three objects share the same acceleration: a = (F-W1-W2-W3) / (m1+m2+m3) = F/(m1+m2+m3) – g = 200/(6+4+5) – 9.8 = 3.53 m/s 2

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Copyright © 2012 Pearson Education Inc. Free-body diagrams Now we focus on m1 and solve for tension T1. T1 W1 for the upper mass, apply the 2 nd law: F-W1-T1 = m1 a we have: T1 = F-W1-m1 a = 200-6*9.8-6*3.53 = 120.02 N

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Copyright © 2012 Pearson Education Inc. Free-body diagrams Similarly we focus on the rope and solve for tension T2. T1 W2 T2 T1-W2-T2 = m2 a we have: T2 = T1-W2-m2 a = 120.02-4*9.8-4*3.53 = 66.7 N

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Copyright © 2012 Pearson Education Inc. Free-body diagrams We can double check the result on the lower mass: T2 W3 T2-W3 = m3 a we have calculated: T2 = 66.7 N a = 3.53 m/s2 66.7 – 5*9.8 = 5 * 3.53? true!

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Copyright © 2012 Pearson Education Inc. Free-body diagrams—examples A free-body diagram is a sketch showing all the forces acting on an object.

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Copyright © 2012 Pearson Education Inc. Free-body diagrams—examples A free-body diagram is a sketch showing all the forces acting on an object.

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Copyright © 2012 Pearson Education Inc. Free-body diagrams—examples

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© 2012 Pearson Education, Inc. Q4.9 A person pulls horizontally on block B, causing both blocks to move horizontally as a unit. There is friction between block B and the horizontal table. If the two blocks are moving to the right at constant velocity, A. the horizontal force that B exerts on A points to the left. B. the horizontal force that B exerts on A points to the right. C. B exerts no horizontal force on A. D. not enough information given to decide

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© 2012 Pearson Education, Inc. A4.9 A person pulls horizontally on block B, causing both blocks to move horizontally as a unit. There is friction between block B and the horizontal table. If the two blocks are moving to the right at constant velocity, A. the horizontal force that B exerts on A points to the left. B. the horizontal force that B exerts on A points to the right. C. B exerts no horizontal force on A. D. not enough information given to decide

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© 2012 Pearson Education, Inc. A woman pulls on a 6.00- kg crate, which in turn is connected to a 4.00-kg crate by a light rope. The light rope remains taut. A. is subjected to the same net force and has the same acceleration. B. is subjected to a smaller net force and has the same acceleration. C. is subjected to the same net force and has a smaller acceleration. D. is subjected to a smaller net force and has a smaller acceleration. E. none of the above Q4.12 Compared to the 6.00-kg crate, the lighter 4.00-kg crate

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© 2012 Pearson Education, Inc. A woman pulls on a 6.00- kg crate, which in turn is connected to a 4.00-kg crate by a light rope. The light rope remains taut. A. is subjected to the same net force and has the same acceleration. B. is subjected to a smaller net force and has the same acceleration. C. is subjected to the same net force and has a smaller acceleration. D. is subjected to a smaller net force and has a smaller acceleration. E. none of the above A4.12 Compared to the 6.00-kg crate, the lighter 4.00-kg crate

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© 2012 Pearson Education, Inc. A car engine is suspended from a chain linked at O to two other chains. Which of the following forces should be included in the free-body diagram for the engine? A. tension T 1 B. tension T 2 C. tension T 3 D. two of the above E. T 1, T 2, and T 3 Q5.1

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© 2012 Pearson Education, Inc. A car engine is suspended from a chain linked at O to two other chains. Which of the following forces should be included in the free-body diagram for the engine? A. tension T 1 B. tension T 2 C. tension T 3 D. two of the above E. T 1, T 2, and T 3 A5.1

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Copyright © 2012 Pearson Education Inc. Two-dimensional equilibrium A car engine hangs from several chains.

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Copyright © 2012 Pearson Education Inc. Two-dimensional equilibrium A car engine hangs from several chains.

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© 2012 Pearson Education, Inc. Q5.3 When released, the cart accelerates up the ramp. Which of the following is a correct free-body diagram for the cart? A.B.C.D. m1am1am1am1a w1w1 w1w1 w1w1 w1w1 TTTT nnnn A cart (weight w 1 ) is attached by a lightweight cable to a bucket (weight w 2 ) as shown. The ramp is frictionless.

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© 2012 Pearson Education, Inc. A5.3 When released, the cart accelerates up the ramp. Which of the following is a correct free-body diagram for the cart? A.B.C.D. m1am1am1am1a w1w1 w1w1 w1w1 w1w1 TTTT nnnn A cart (weight w 1 ) is attached by a lightweight cable to a bucket (weight w 2 ) as shown. The ramp is frictionless.

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© 2012 Pearson Education, Inc. Q5.4 A. T = w 2 B. T > w 2 C. T < w 2 D. not enough information given to decide When released, the cart accelerates up the ramp and the bucket accelerates downward. How does the cable tension T compare to w 2 ? A cart (weight w 1 ) is attached by a lightweight cable to a bucket (weight w 2 ) as shown. The ramp is frictionless. The pulley is frictionless and does not rotate.

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© 2012 Pearson Education, Inc. A5.4 A. T = w 2 B. T > w 2 C. T < w 2 D. not enough information given to decide When released, the cart accelerates up the ramp and the bucket accelerates downward. How does the cable tension T compare to w 2 ? A cart (weight w 1 ) is attached by a lightweight cable to a bucket (weight w 2 ) as shown. The ramp is frictionless. The pulley is frictionless and does not rotate.

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Copyright © 2012 Pearson Education Inc. Bodies connected by a cable and pulley If the cart is moving at constant speed, what is w2/w1?

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Copyright © 2012 Pearson Education Inc. Bodies connected by a cable and pulley If the cart is moving at constant speed, what is w2/w1? consider the bucket T = w2

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Copyright © 2012 Pearson Education Inc. Bodies connected by a cable and pulley If the cart is moving at constant speed, what is w2/w1? consider the cart along x direction: W1 Sin 15 o – T = 0 Thus: W2 = W1 Sin 15 o and: W2/W1 = Sin 15 o = 0.26

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