A 41.8 kg box rests on a ramp with a slope of 60 degrees. Find the force of friction acting on the box.

Answer 1

Draw out your diagram so you can visualize the answer and label the sum of the forces.

The equation for static frictiion --> F_s≤N

F_f = μ_sN = μ_smgcosθ

Because this object is in equilibrium the sum of the forces is zero.

x-direction ∑F = mgsinθ - F_f = 0

μ is unknown therefore, we solve it first.

Therefore,

mgsinθ - μ_smgcosθ = 0, so doing some algebra we get:

μ_s = (mgsinθ)/(mgcosθ) = tanθ = tan60 = √3

Now we can solve the question:

F_f = μ_sN = μ_smgcos60 = -(√3 • 41.8 • -10 • 1/2) =362 N

So, this is about what I got. My professor was cool and he wanted us to use -10 m/s² for g. Moreover, check your sig figs cause I always lost points for them. I am lazy when it comes to them. Plus, double check my work

Answer 2

Hadassah I wish I was able to draw here. But the key is that this object is at rest. Therefore, this is known as static friction and means the friction is less than or close to N (the normal force).

Therefore, the equation is rather simple: the sums of the forces (ΣF = mā) are

F_f = mgsinθ

Thus, the force of friction is equal to the pull of the force of gravity.

So, plug and chug