Question

Call the stack of two bricks system A and the single brick System B A Constant speed

Part 3:

Let system A now contain just the lower brick on the stack of two bricks. Call the top brick system C. Now we have three systems, one for each brick. The hand is pushing horizontally with a constant speed. In this situation the coefficient of friction between the bricks and table is less than that in question 1 (where the friction was not negligible and the coefficient of friction was not given).

A. Compare the magnitude of the net force on System C to the magnitudes of the net forces on systems A and B. Explain.

B. Draw a free body diagram for each of the three systems. List any pairs of forces that are of equal magnitude and explain why for each pair.

C. Now define yet another system, D, that contains all three bricks. Compare the magnitude of the net force on system D to the magnitudes of the net forces on systems A, B and C, as well as on the original system A of two bricks.

D. Draw a free body diagram for system D.

1. Are there any forces on your diagrams for systems A, B and C in part A that are not on your free body diagram for system D? If so, what characteristic do these forces have in common that none of the other forces share?

2. Why is it not necessary to consider these forces in determining the motion of system D?

Answer 1

Since acceleration system A, B, C is zero as moving with contain speed. Thus net force of system A, B and C is zero. Thw for system A, F_1 = F_2 + 2lmg = 3lmg No force in acting as is any force acts it will have acceleration. As explained in case 1, since body is moving with constant speed so any system we consider it must have net force zero. FBD of system D Yes, Reaction force on B(ie F_2)since in system D it because internal force. By Newton's 3^rd law every action equal to opposite reaction so these internal forces will cancel each other, thus they do not affect motion of body.