# Hanging mass force

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 wouldnt think so because we are trying to figure out how much mass is need to start the first mass moving, so the tension doesnt 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

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