Problem: Two masses are connected by a cord and placed on a cart as shown. What must the acceleration of the cart be (caused by the force F) so that the mass on the right does not move relative to the vertical surface of the cart? What is the tension in the cord? What must the force F be? Both surfaces have a static frictional coefficient of 0.76 and a kinetic coefficient of 0.33. M = 12 kg, m₁ = 4 kg, and m₂ = 2.5 kg.

Solution: As before, we draw the free body diagrams. Since m₁ tends to be dragged to the right by the cart, the frictional force on it must point to the right. Since m₂ tends to fall, its frictional force points up. Both are static frictional forces, since the bodies do not move relative to each other. Newton's Third Law (action-reaction) then demands that the reactions to these frictional forces (on the contacting surfaces) must point oppositely:

Writing Newton's Second Law for body 2:

Writing Newton's Second Law for body 1:

Subtracting these two equations,

To get the force F, the free body of the cart is necessary (the only place F appears):

Is a smaller acceleration possible? If there were less acceleration, is it possible that the 4 kg block (on top) would tend to slide forward because of the hanging block? That would mean changing the directions of f₁ on the 12 kg mass and the 4 kg mass, and redoing the problem. Something for you to try...