Question

You are lowering two boxes, one on top of the other, down the ramp shown in the figure by pulling on a rope parallel to the surface of the ramp. Both boxes movetogether at a constant speed of 17.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.499, and the coefficient of static frictionbetween the two boxes is 0.822.

What force do you need to exert to accomplish this?

What is the magnitude of the friction force on the upper box?

Answer 1

hypotenuse = sqrt ( 2.50² + 4.75² ) = 6.25 + 22.56 = 28.81 m

sinθ = 2.50 / 28.81 = 0.0015
frictional force on upper box = fr =μ_smg sinθ = 0.822 * (32) * 9.8 * 0.0015
= 0.3866 N


for a = 0 , force exerted by me on the box should be balanced by the frictional force acting on lower box in the front direction

=> F = μ_kR = μ_k (m1+m2)g sinθ = 0.499 * (32 + 48) * 9.8 * 0.0015 = 0.5868 N

Answer 2

For part a:

I'm going to make @ = theta because I feel like it

T = (m1+m2)*g*(sin@ - 0.473cos@)

You'll find @ by using the pythagrean theorem (a^2+b^2=c^2), then basic soh-cah-toa trig stuff

However I'm still trying to figure out part b

good luck :)

Answer 3

The frictional force opposing the rope will be
F_f=μmgsinθ

F_f=.499(32+48)(9.8)sinθwhereθ=arctan(2.5/4.75)

F_f=72.883

In order to have a constant velocity, the applied force be equivalent to this frictional force so:

F=72.883N

The frictional force on the upper box is the same equation above expect the mass term is just 32, not the sum of both as it was in the equation above

F_f=.499(32)(9.8)sinθ

F_f=29.153 N

hope that helps