A box (m = 20 kg) is sitting on a horizontal surface. It is connected to a massless hook by a light string passing over a massless pulley wheel. The coefficients of friction between the box and the surface are 0.50 (static) and 0.30 (kinetic). On top of m is a second box M of mass 20 kg. The coefficients of friction between the boxes m and M are 0.80 (static) and 0.60 (kinetic). How much weight needs to be added to the hook until the box just begins to move.
A) 10 kg
B) 98 kg
C) 20 kg
D) 16 kg
E) 200 kg

Question

A box (m = 20 kg) is sitting on a horizontal surface. It is  connected to a massless hook by a light string passing over a massless pulley wheel. The coefficients of friction between the box and the surface are 0.50 (static) and 0.30 (kinetic). On top of m is a second box M of mass 20 kg. The coefficients of friction between the boxes m and M are 0.80 (static) and 0.60 (kinetic). How much weight needs to be added to the hook until the box just begins to move. 
A) 10 kg 
B) 98 kg 
C) 20 kg 
D) 16 kg 
E) 200 kg

Quizplus · Accepted Answer

The force acting on the system is the weight of the added weight (W) minus the weight of the two boxes (40 kg) times the acceleration of the system (a). Since the boxes are not moving, the acceleration is zero, so we have:

W - 40g * 0 = 0

W = 40g

W = 40 * 9.8 ≈ 392 N

To overcome static friction between the box and the surface, the force applied by the string (T) must be greater than or equal to the maximum static friction force (Fsmax = μsN). The normal force (N) acting on the box is equal to the weight of the box (mg). Thus, we have:

T - μsN = 0.5 * 20 * 9.8

T - 98 = 98

T = 196 N

The force applied by the string must also be equal to or greater than the weight of the added weight (W) for the system to begin moving. Thus, we have:

T - W = 0

T = W

W = 196 N

The weight that needs to be added to the hook is:

W/g = 196/9.8 ≈ 20 kg.

A Box (M = 20 Kg) Is Sitting on a Horizontal