Question

A 100-kg box is placed on a ramp. As one end of the ramp is raised, the box begins to move downward just as the angle of inclination reaches 25°. What is the coefficient of static friction between box and ramp?

Answer 1

Answer:

$\mu_s=0.47$

Explanation:

Just before the box starts moving, according to Newton's first law, we have:

$\sum F_x:F_f-W_x=0(1)\\\sum F_y:N-W_y=0(2)$

The sine of the angle (25°) of a right triangle is defined as the opposite cathetus ($W_x$) to that angle divided into the hypotenuse (W):

$sin25^\circ=\frac{W_x}{W}\\W_x=Wsin25^\circ\\W_x=mgsin25^\circ(3)$

The cosine of the angle (25°) of a right triangle is defined as the adjacent cathetus ($W_y$) to that angle divided into the hypotenuse (W):

$cos25^\circ=\frac{W_y}{W}\\W_y=Wcos25^\circ\\W_y=mgcos25^\circ(4)$

Recall that the maximum frictional force is defined as:

$F_f=\mu_s N(5)$

Replacing (3) in (1) and (4) in (2):

$F_f=mgsin25^\circ(6)\\N=mgcos25^\circ(7)$

Replacing (7) and (6) in (5) and solvinf for $\mu_s$:

$\mu_s=\frac{mgsin25^\circ}{mgcos25^\circ}\\\mu_s=\frac{sin25^\circ}{cos25^\circ}\\\mu_s=0.47$