Question

A JNT delivery man is delivering a big box to a client. He is dragging it up a concrete slope leading to the porch, with an angle of 15° to the horizontal. Then he realized that he is on the wrong location so he again dragged the box down the concrete slope. To get the box started up the slope, it is necessary that the man exert six times the force needed to get it started down the slope. If the force is always parallel to the slope, what is the coefficient of static friction between the box and the concrete?

Answer 1

Explanation & Calculation

• For the movement up the incline,

$\qquad\qquad \begin{aligned} \small F&=\small mg\sin\theta+\mu R=mg\sin\theta+\mu mg\cos\theta\\ \end{aligned}$

• For that down the incline,

$\qquad\qquad \begin{aligned} \small \frac{F}{6}&=\small \mu R-mg\sin\theta = \mu mg\cos\theta-mg\sin\theta \end{aligned}$

• Upon the replacement of $\small F$

$\qquad\qquad \begin{aligned} \small \sin\theta+\mu\cos\theta&=\small 6(\mu\cos\theta-\sin\theta)\\ \small \mu&=\small \frac{7}{5}\tan\theta=\frac{7}{5}\tan(15)\\ &=\small 0.38 \end{aligned}$