Question

In the following picture, the slope has a static coefficient of friction of .5 and a kinetic coefficient of .4. The incline has a 42 degree angle.

a) In the absence of friction, which way would the boxes move? (Justify with numbers)

b) Determine if the boxes will move, given that there is friction.

c) If the boxes are initially in motion, determine the acceleration of the system

Answer 1

I believe that the system will move to the right since the 8kg box weighs more than the 5kg box by using weight(w)= mass*gravity for both boxes, is that correct?

But I do not know how to begin solving part b or c, so I would be very grateful for any help given!

Answer 2

you should assume that the boxes will move on to one direction with an acceleration (say a). since both boxes should move with the same acceleration, you can apply F=ma to find the acceleration.

But if you want to get an approximate answer, you should get the weight component of the 8kg along the direction of the slope(mg sin(theta)). That is the force which affects the tension of the rope not the whole mg force

Answer 3

Alright, that makes sense since the incline has to affect the way gravity is acting on the 8kg box. So using that equation mg sin(theta), the weight for the 8kg box still turns out to be greater than the hanging 5kg box, so the system should still move to the right.

How do we know that the boxes even move in the first place? Is it because mu kinetic is smaller than mu static?

Answer 4

\[F_1 = 5g\] and \[F_2 = (8\times \sin42^\circ ) g \approx 5.35g \]
so with no friction the 8 kg weight pulls up the 5 kg weight.

Now if there is friction and the system is at rest initially then

\[F_1 = 5g\] and \[F_2 = (8\times \sin42^\circ ) g \approx 5.35g \]
so with no friction the 8 kg weight pulls up the 5 kg weight.

Now if there is friction it is
\[\mu N = 0.5 \times (8\times \cos42^\circ )g \approx 2.97g\]
and \[F_1 + \mu N = 5g+2.97g=7.97g >5.35g\]

This also applies to the case when they move
\[\mu N = 0.5 \times (8\times \cos42^\circ )g \approx 2.97g\]
and \[F_1 + \mu N = 5g+2.97g=7.97g >5.35g\]

Now if there is friction it is
\[\mu N = 0.4 \times (8\times \cos42^\circ )g \approx 2.38g\]
and \[F_1 + \mu N = 5g+2.38g=7.38g >5.38g\]