Forces of Friction | Application of Newton's Laws

Question

Question: In Figure P.36, Mass of 1 = 10kg and Mass of 2 equals 4.0 kg. The coefficient of static friction between mass of 1 and the horizontal surface is .50, and the coefficient of kinetic friction is .30. (a) If the system is released from rest, what will its acceleration be? (b) If the system is set in motion with mass of 2 moving downward, what will be the acceleration of the system?

I am having trouble to conceptually solve this problem. I drew both FBD's on my paper, and see that mass of 1 has three forces (F normal, F tension, and F gravity). Mass of 2 has three forces acting on it (F tension, and F gravity). I do not know even know if that is right. For part a, will acceleration be zero since mass of two is less than mass of one, not being able to accelerate? Is there something else I am missing? How do you solve part b?

Answer 1

The normal force due to gravity is given by F_n = 10 kg x 9.8 m/s² = 98 newtons. The force of static friction is

F_s = 49 x 0.5 = 49 newtons. This is a retarding force which acts in the direction opposite to the force exerted by block M₂ which is 4 kg x 9.8 m/s² = 39.2 newtons. As the force tending to move the mass M₁

( 39.2 newtons) is less than the retarding force (49 newtons), the block won't move so the acceleration is zero,

Once the block is moving, the retarding force decrease to 29.4 newtons (0.3 x 10 x 9.8 = 29.4. Now the net force of 9.8 newtons (39.2-26.4) can be set equal to the mass of block M₁ x a and solved for a. I got 0.98 m/s². The key is to remember that the force due to friction retards motion.

Answer 2

You're on the right track, but missing one important force: friction. Friction is going to be the force that either stops or reduces our acceleration. It always acts in the opposite direction of the motion, or potential motion, of the object. Here, we should add friction as a force on the free body diagram for mass 1, going to the left.

Then, in order to determine whether our system overcomes friction, we need the formula. This is f_s,k = μ_s,kN, where f is the friction force, μ (greek mu) is the coefficient of friction, N is the normal force, and the subscripts s and k refer to static (stationary) or kinetic (moving) friction. In order to get moving in part a) we would need to overcome the force of friction. The force opposing friction is tension, so we need to determine if tension is equal to or greater than the force of tension.

For part b) we are already in motion, so we would set the problem up very similarly, just using the coefficient of kinetic friction instead of static friction. From there we can determine the acceleration normally according to Newton's 2nd law.