Question

You are lowering two boxes, one on top of the other, down a ramp by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800.


(a) What force do you need to exert to accomplish this? (b) What are the magnitude and direc- tion of the friction force on the upper box?

Answer 1

Answer:

(a) 57.17 N

(b) 146.21 N up the ramp

Explanation:

Assume the figure attached

(a)

The angle of ramp, $\theta=tan ^{-1} \frac {2.5}{4.75}=27.75854^{\circ}\approx 27.76^{\circ}$

$N=(32+48)*9.81*cos 27.76^{\circ}=694.4838 N$

m=32+48=80 kg

$T+ \mu N= mg sin \theta$

$T=mg sin \theta - \mu N= mg sin \theta- \mu mg cos \theta= mg (sin \theta - \mu cos \theta)$ where $\mu$ is coefficient of kinetic friction

$T=80*9.81(sin 27.76^{\circ} -0.444 cos 27.76^{\circ})=57.16698 N \approx 57.17 N$

(b)

Upper box doesn’t accelerate

$F_r= mgsin\theta= 32*9.81sin 27.76^{\circ}=146.2071\approx 146.21 N$  

The direction will be up the ramp