Question

A box is dragged across a horizontal floor at constant velocity by means of a rope tied to the front of the box. The rope makes an angle of 12° with the horizontal (above the horizontal). The magnitude of the force applied to the rope is half the magnitude of the earth’s gravitational force on the box. Find the coefficient of friction governing the friction between the box and the floor. Write your answer to two significant figures.

I dont understand how to solve it without the mass of the box...

Answer 1

**The magnitude of the force applied to the rope is half the magnitude of the earth’s gravitational force on the box**

(1) |F| = 0.5 mg
(2) |F|cos theta = ma - R*mu = 0
(3) R = mg - |F|sin theta