Force Required to Move Box of Mass 20 kg w/ Friction

[StephenDoty]

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


So the total mass of m= 40kg
so fs=40*9.80*.5
so fs= 196N
thus the force pulling down on the second mass has to equal fs
the force pulling down is mg
so mg=196N
m=20kg

I do realize that on the second mass hanging over the table has both a tension force and a mg force, do you have to take the tension into account? I wouldn't think so because we are trying to figure out how much mass is need to start the first mass moving, so the tension doesn't matter, its just the force that carries the mg force from the second mass to the first mass. Making mg=fs

Isnt the force of the system = 0 before it starts moving thus 0= mg -fs or mg=fs?
I do not like my way of solving this problem. How would you guys solve it??

Thanks
Stephen

 

[Nate810]

Yes, your approach is correct. You can solve this problem by using the equation Fnet = ma, where Fnet is the net force acting on the box m, m is the mass of the box, and a is the acceleration of the box (which is zero since it is not moving). This equation can be rearranged to solve for Fnet: Fnet = ma = m*0 = 0. Since the net force is equal to zero, all the forces acting on the box must be balanced. The two forces acting on the box are the force due to gravity (mg) and the force of friction (fs). Therefore, mg = fs. To find the amount of weight that needs to be added to the hook, you need to calculate the force of friction. The force of friction is given by fs = μs*N, where μs is the coefficient of static friction between the box and the surface, and N is the normal force. The normal force is equal to mg, so fs = μs*mg = μs*m*g. Thus, the amount of weight that needs to be added to the hook until the box just begins to move is: W = μs*m*g = 0.5*20*9.8 = 98 kg.

 

[Moetasim]

I would approach this problem by first identifying the forces acting on the system and then applying Newton's laws of motion to determine the force required to move the box.

In this scenario, there are several forces at play: the weight of the box (mg), the tension in the string (T), and the force of friction (Ff). The tension and friction forces will vary depending on whether the box is at rest or in motion.

To determine the force required to move the box, we need to consider the maximum force of friction that can be exerted on the box without causing it to move. This maximum force of friction is known as the limiting friction force (Ffmax) and can be calculated using the coefficient of friction and the weight of the box.

For the first box (m), the maximum force of friction (Ffmax) can be calculated as follows:

Ffmax = μs * mg = 0.50 * 20kg * 9.8m/s^2 = 98N

Similarly, for the second box (M), the maximum force of friction (Ffmax) can be calculated as follows:

Ffmax = μs * Mg = 0.80 * 20kg * 9.8m/s^2 = 156.8N

Since the two boxes are connected by a string, the tension force (T) must be the same for both boxes. Therefore, in order for the first box to just begin to move, the tension force (T) must be equal to the maximum force of friction (Ffmax) acting on it.

Thus, we can set up the equation T = Ffmax and solve for T:

T = Ffmax = 98N

Therefore, the weight that needs to be added to the hook is equal to the tension force (T), which is 98N.

In conclusion, the correct answer is A) 10 kg.