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Question

A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 29 ∘ above the horizontal. The coefficients of static and kinetic friction between the box and wall are 0.40 and 0.30, respectively. The box slides down unless the applied force has magnitude 33 N .

Part A What is the mass of the box? Express your answer using two significant figures.

Answer 1

Draw a sketch and a free-body diagram:

If F = 33, then that is the equilibrium point where no motion is occurring but if slightly less force is applied, the object breaks free and starts sliding down the wall.

We can break F into its x-component and y-component. I'll set my coordinate plane up so the origin is at the "dot" in the free-body diagram and up is positive "y" and right is positive "x".

Fx = Fcos(29°) = 33cos(29°) = 28.8625 N

Fy = Fsin(29°) = 33sin(29°) = 15.9987 N

Remember that F_F = μF_N (or F_N = F_F/μ)but in this case, we can think of μ as the static coefficient of friction since the object is not moving.

Also remember the F_G = mg (the weight of the object)

In the x-direction:

∑F_x = 0 = F_x - F_N = 28,28625 - F_F/μ = 28,28625 - F_F/0.4

So 28,28625 = F_F/0.4 or F_F = 11.545 N

In the y-direction:

∑Fy = 0 = F_F + F_y - F_G = 11.545 + 15.9987 - m(9.81)

So 11.545 + 15.9987 = 9.81m

27.5437/9.81 = m = 2.8 kg