Question

A block with an unknown mass is being pushed on by a force equal to F = 260 N, applied to the box at an angle 20 degrees from horizontal. There is significant friction between the block and the surface, which prevents the box from sliding.

The coefficient of static friction between the box and the surface is most nearly…

(A) 0.34

(B) 0.55

(C) 0.94

(D) It cannot be determined

Answer 1

Explanations & Calculations

• To calculate the coefficient of static friction, the given situation should correspond to the static equilibrium in which static friction equals the applied force but we are clearly told
• Only the horizontal component of the applied force: $\small F\cos20$ contributes to the motion of the box while the vertical component: $\small F\sin20$ works in increasing the reaction force on the box from the ground.
• If we wrote equations to assess the static friction (assuming static equilibrium is achieved) we will see,

$\qquad\qquad \begin{aligned} \small F\cos20&=\small \mu_sR\\ \small F\cos20&=\small \mu_s(mg+F\sin20)\\ \small \mu_s&=\small \frac{F\cos20}{mg+F\sin20} \end{aligned}$

• This concludes that the needed quantity cannot be assessed with only the given information.