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

First, let's analyze the forces acting on box M:

The weight of M is 20g = 196 N (where g is the acceleration due to gravity).

The normal force on M is equal to its weight, since it is resting on a horizontal surface.

The frictional force between m and M is 0.8 times the normal force, or 0.8 * 196 = 157 N.

Next, let's analyze the forces acting on box m:

The weight of m is also 20g = 196 N.

The normal force on m is equal to its weight plus the weight of M (since M is sitting on top of m), or 2 * 196 = 392 N.

The frictional force between m and the surface is either a static frictional force of 0.5 * 392 = 196 N (if the box is at rest) or a kinetic frictional force of 0.3 * 392 = 118 N (if the box is in motion).

Now, we can analyze the forces on the system as a whole (m and M combined):

The weight of the system is 392 N + 196 N = 588 N.

The normal force on the system is equal to its weight, since it is resting on a horizontal surface.

The frictional force between the boxes is either a static frictional force of 0.8 times the normal force, or 0.8 * 588 = 470 N (if the boxes are at rest) or a kinetic frictional force of 0.6 times the normal force, or 0.6 * 588 = 353 N (if the boxes are in motion).

The tension in the string is the same throughout the system (since it is a light string with no mass), and is acting upwards.

To find the weight that needs to be added to the hook until the system just begins to move, we need to find the minimum tension in the string required to overcome the frictional forces acting on the system.

If the system is at rest, then the tension in the string must be equal to the static frictional force between the system and the surface, which is 470 N.

If the system is in motion, then the tension in the string must be equal to the kinetic frictional force between the system and the surface, which is 353 N.

So, in order to find the weight that needs to be added to the hook until the system just begins to move, we need to compare these two values and see which one is larger.

Since 470 N is greater than 353 N, we know that the system is at rest and the tension in the string is 470 N.

The weight that needs to be added to the hook is equal to the difference between the weight of the system and the tension in the string, or 588 N - 470 N = 118 N.

Since the weight of each block is 20g = 196 N, we need to add a weight of 118/196 = 0.6 kg to the hook.

Therefore, the answer is C) 20 kg.

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