Worksheet #04 Newton s Laws of Motion & Free-Body-Diagram.

Transcription

1 Worksheet #04 1. Three identical blocks are pushed across a table at constant speed as shown below. The hand pushes horizontally. There is friction between the table and the bricks. We shall call the stack of two bricks system A and the single brick, system B. A B (a) Compare the net force (magnitude and direction) on systems A and B. (b) Draw the free body diagram for each system in the spaces below. Clearly label all forces (eg: a normal force on A by the hand should be labeled N A,hand or something of the sort). System A System B (c) Consider the following discussion between two students: Student 1: I think the free-body diagram for system B should have a force by A and a force by the hand. Student 2: I don t think the diagram should include a force by the hand. People cannot exert forces on objects without touching them. With which student, if either, do you agree? Explain your reasoning. If you disagree with a statement, explain what is wrong with it. (d) There are many types of forces and they can always be divided into two categories: Contact forces and non-contact forces. Categorize the forces you used in the free-body diagrams. Where would you include tension and magnetic forces? 1

2 Worksheet #04 Contact forces: Non-contact forces: (e) Compare the force (magnitude and direction) done on system A by system B to the force done on B by A. (f) Identify all Newton s third law pairs in your diagrams by placing a small "x" symbol on each member of the pair (use "xx" for the second pair, etc) or color code them if possible. (g) Write the Newton s second law equations for each system. equations per system). (Hint: There should be two (h) Suppose the mass of each brick is m = 2.5 kg, the friction on system A is 10 N, the friction on system B is 5.0 N, and the bricks are moving at a constant speed of 0.50 m/s. Determine the magnitude of each of the forces in your diagrams. (Use g = 10 m/s 2 ) (i) Would your answers change if the bricks were moving twice as fast? 2

3 Worksheet #04 2. The blocks are now pushed by the hand with the same force as in problem 1, but the friction is now less than before. A B (a) Describe the motion of the bricks. (b) Compare the net force (magnitude and direction) on systems A and B. (c) Draw the free body diagram for each system in the spaces below. Clearly label all forces (eg: a normal force on A by the hand should be labeled N A,hand or something of the sort). System A System B (d) Consider the following discussion between two students: Student 1: The system is pushed by the same force as before, so they will have the same motion as before. Student 2: I disagree. I think they are speeding up since friction is less. Now system A is pushing on system B with a greater force than system B is pushing on A. With which student, if either, do you agree? Explain your reasoning. If you disagree with a statement, explain what is wrong with it. 3

4 Worksheet #04 3. Let C be the system containing all three bricks. The motion of the bricks is the same as in question 2 C (a) Compare the net force on systems A and B (as discussed in question 2) with C. (b) Draw the free body diagram for the system in the space below. Clearly label all forces (eg: a normal force on A by the hand should be labeled N A,hand or something of the sort). System C (c) Assume m = 2.5 kg, and that the friction is 5.0 N for system A and 2.5 N for system B. Find the acceleration of the system and the magnitude of the normal force on system A by system B. 4

5 Worksheet #04 4. Two blocks of masses m 1 = 5.0 kg and m 2 = 2.0 kg hang on both sides of an incline, connected through an ideal, massless string that goes through an ideal, massless pulley, as shown in figure. The friction between the blocks and the incline is negligible. The angle between the incline and the horizontal is θ = 30. (a) Find the acceleration of mass 2. m 1 m 2 θ (b) Determine the magnitude of the tension in the string. (c) If mass 1 starts moving at h = 1.5 m from the ground, with what velocity will it hits the ground? 5

6 Worksheet #04 5. A 30 mg-spider hangs from two threads from the ceiling as shown in the figure. (a) Find the tension on each thread ::: (b) The ceiling happens to be the ceiling of an elevator that starts moving up with an acceleration of 2.5 m/s 2. Find the new tensions on the threads. 6

7 Solution for Worksheet #04 1. Three identical blocks are pushed across a table at constant speed as shown below. The hand pushes horizontally. There is friction between the table and the bricks. We shall call the stack of two bricks system A and the single brick, system B. A B (a) Compare the net force (magnitude and direction) on systems A and B. Answer: The net force is zero in both cases because the speed is constant (i.e. a = 0) (b) Draw the free body diagram for each system in the spaces below. Clearly label all forces (eg: a normal force on A by the hand should be labeled N A,hand or something of the sort). System A N A,table System B N B,table N A,B f A,table X N A,hand f B,table X NB,A W A,Earth W B,Earth Answer: See diagrams above. (vector lengths not to scale) (c) Consider the following discussion between two students: Student 1: I think the free-body diagram for system B should have a force by A and a force by the hand. Student 2: I don t think the diagram should include a force by the hand. People cannot exert forces on objects without touching them. With which student, if either, do you agree? Explain your reasoning. If you disagree with a statement, explain what is wrong with it. Answer: Student 2 is right. The ultimate reason why A is pushing on B is the hand, but that does not mean that the force by the hand acts on B. The force on B is exerted by A (N B,A ) and is different from the force exerted on A by the hand (N A,hand ). (d) There are many types of forces and they can always be divided into two categories: Contact forces and non-contact forces. Categorize the forces you used in the free-body diagrams. Where would you include tension and magnetic forces? Contact forces: Non-contact forces: Answers: Normal Weight (gravitational) Friction Magnetic Tension Electric 1

8 Solution for Worksheet #04 (e) Compare the force (magnitude and direction) done on system A by system B to the force done on B by A. Answer: They have equal magnitude and opposite direction. (Newton s third law) (f) Identify all Newton s third law pairs in your diagrams by placing a small "x" symbol on each member of the pair (use "xx" for the second pair, etc) or color code them if possible. Answer: See figure. Newton pairs have non-black identical colors. Only one pair is identified in systems drawn above which draw on red (also marked by "x" for the color blind). (g) Write the Newton s second law equations for each system. equations per system). Answer: System A N A,hand N A,B f A,table = m A a (Hint: There should be two System B N B,A f B,table = m B a N A,table W A,Earth = 0 N B,table W B,Earth = 0 (h) Suppose the mass of each brick is m = 2.5 kg, the friction on system A is 10 N, the friction on system B is 5.0 N, and the bricks are moving at a constant speed of 0.50 m/s. Determine the magnitude of each of the forces in your diagrams. (Use g = 10 m/s 2 ) Answer: Constant speed means a = 0. Substituting the quantities given above, the system of equations now reads: N A,hand N A,B 10 N = 0 [1] N B,A 5.0 N = 0 [3] N A,table (5.0 kg) ( 10 m/s 2) = 0 [2] N B,table (2.5 kg) ( 10 m/s 2) = 0 [4] Equation [3] gives: N A,B = N B,A = 5.0 N. N B,A = 5.0 N, but Newton s 3rd law requires N A,B = N B,A ; therefore, Equation [2] gives: N A,table = 50 N. and equation [4] gives: N B,table = 25 N. finally, equation [1] gives: N A,hand = 5.0 N + 10 N = 15 N (i) Would your answers change if the bricks were moving twice as fast? Answer: No. The speed does not matter as long as it is constant. 2

9 Solution for Worksheet #04 2. The blocks are now pushed by the hand with the same force as in problem 1, but the friction is now less than before. A B (a) Describe the motion of the bricks. Answer: If friction is less than before and the pushing force is the same, now the system will have a net force to the right and thus an acceleration to the right. The blocks move faster and faster. (b) Compare the net force (magnitude and direction) on systems A and B. Answer: The acceleration is the same for all three systems, so the net force is proportional to the mass. F net,a = 2F net,b (c) Draw the free body diagram for each system in the spaces below. Clearly label all forces (eg: a normal force on A by the hand should be labeled N A,hand or something of the sort). System A N A,table System B N B,table N A,B X f A,table N A,hand f B,table X NB,A W A,Earth W B,Earth Answer: See diagrams above. (vector lengths not to scale.) Note that FBD did not change despite the fact the system now has an acceleration. (d) Consider the following discussion between two students: Student 1: The system is pushed by the same force as before, so they will have the same motion as before. Student 2: I disagree. I think they are speeding up since friction is less. Now system A is pushing on system B with a greater force than system B is pushing on A. With which student, if either, do you agree? Explain your reasoning. If you disagree with a statement, explain what is wrong with it. Answer: Student 1 is forgetting that it is the sum of all forces that determines the acceleration and therefore the motion. One force being the same is not enough to guarantee the same motion. Student 2 gets the first sentence right, but the second part is wrong. System A pushed on B with a force of exactly the same magnitude as system B pushes on A because of the third law. 3

10 Solution for Worksheet #04 3. Let C be the system containing all three bricks. The motion of the bricks is the same as in question 2 C (a) Compare the net force on systems A and B (as discussed in question 2) with C. Answer: The acceleration is the same for all three systems, so the net force is proportional to the mass. F net,b = 1 2 F net,a = 1 3 F net,c (b) Draw the free body diagram for the system in the space below. Clearly label all forces (eg: a normal force on A by the hand should be labeled N A,hand or something of the sort). System C N C,table N C,hand f C,table W C,Earth (c) Assume m = 2.5 kg, and that the friction is 5.0 N for system A and 2.5 N for system B. Find the acceleration of the system and the magnitude of the normal force on system A by system B. Answer: Newton s 2nd law for system C, in the horizontal direction can be obtained from the FBD on system C: N C,hand f C,table = m C a Using N hand,c = 15 N as found in question 2 and the new quantities given for the friction forces, the last equation reads: 15 N 7.5 N = (7.5 kg) a a = 1.0 m/s 2 To find the magnitude of the magnitude of the normal force on system A by system B, we need to return to the FBD drawn in question 2 for system B. N B,A f B,table = m B a N B,A 2.5 N = (2.5 kg) ( 1.0 m/s 2) N B,A = 5.0 N 4

11 Solution for Worksheet #04 4. Two blocks of masses m 1 = 5.0 kg and m 2 = 2.0 kg hang on both sides of an incline, connected through an ideal, massless string that goes through an ideal, massless pulley, as shown in figure. The friction between the blocks and the incline is negligible. The angle between the incline and the horizontal is θ = 30. (a) Find the acceleration of mass 2. Answer: Block 1 : T m 1 g = m 1 a (1) Block 2 : m 2 g sin θ T = m 2 a (2) That is a system of two equations with two unknowns a and T. Adding the two equations we get T eliminated, thus: m 1 T m 1 g T m 2 g N m 2 θ y x m 2 g sin θ m 1 g = (m 1 + m 2 ) a (3) a = m 2 sin θ m 1 m 1 + m 2 g (4) = (5.0 kg) sin (30 ) 2.0 kg 5.0 kg kg ( 9.8 m/s 2 ) = 5.6 m/s 2 (5) The acceleration is negative, meaning mass 1 will accelerate downwards and mass 2 will accelerate up the incline. (b) Determine the magnitude of the tension in the string. Answer: If substitute a from equation 4 into the result of equation 1 we get: T = m 1 (g + a) = (5.0 kg) ( 5.6 m/s m/s 2) = 21 N (6) (c) If mass 1 starts moving at h = 1.5 m from the ground, with what velocity will it hits the ground? Answer: This is 1D constant accelerated motion: v 2 = v a x v = ± v a x = ± ( 5.6 m/s 2) ( 1.5 m) Since the block is moving downward, we pick the negative solution (note, the problem asks for velocity not just the speed). because block was in the negative direction of our coordinate system shown in the figure above. v = 4.1 m/s 5

12 Solution for Worksheet #04 5. A 30 mg-spider hangs from two threads from the ceiling as shown in the figure. (a) Find the tension on each thread. Answer: x : T 2 cos θ 2 T 1 cos θ 1 = 0 y : T 2 sin θ 2 + T 1 sin θ 1 mg = 0 θ 1 = 40 T 1 T 2 θ 2 = 30 cos θ 1 T 1 = T 2 cos θ 2 (7) and T 2 sin θ 2 + T 1 sin θ 1 = mg (8) Substitute eq. 7 into eq. 8 and rearrange to get: T 1 cos θ 1 cos θ 2 sin θ 2 + T 1 sin θ 1 = mg ::: mg T 1 cos θ 1 sin θ 2 + sin θ 1 cos θ 2 cos θ 2 = mg (9) T 1 = mg cos θ 2 sin (θ 1 + θ 2 ) = 0.27 mn (10) Where the fact sin(a + B) = sin A cos B + cos A sin B is used. Similarly, we can find the other tension: T 2 = mg cos θ 1 sin (θ 1 + θ 2 ) = 0.24 mn (11) (b) The ceiling happens to be the ceiling of an elevator that starts moving up with an acceleration of 2.5 m/s 2. Find the new tensions on the threads. Answer: The free-body diagram is the same, but now the acceleration in the vertical direction is not zero. The equations for both x and y components will now read: x : T 2 cos θ 2 T 1 cos θ 1 = 0 y : T 2 sin θ 2 + T 1 sin θ 1 mg = ma cos θ 1 T 1 = T 2 cos θ 2 (12) and T 2 sin θ 2 + T 1 sin θ 1 = m(g + a) (13) Equations 12 and 13 are similar to equations 7 and 8 but for the replacement of g with a + g. We expect the tension formulas to be the same as equations 10 and 11 except for replacing g with a + g. T 1 = m (a + g) cos θ 2 sin (θ 1 + θ 2 ) T 2 = m (a + g) cos θ 1 sin (θ 1 + θ 2 ) = 0.34 mn = 0.30 mn 6