Answer to Question #137109 in Mechanics | Relativity for Subhasmita Dash

Question

A crate of mass 75 kg is pulled along a
horizontal floor with constant velocity
by a rope attached to the crate. The
coefficient of kinetic friction between
the floor and the crate is 0.2 and the
angle the rope makes with the
horizontal is 45°. Calculate the
magnitude of the force exerted on the
crate by the rope. Draw the free body
diagram. Take g = 10 ms 2.

Answer 1

The gravitational force is $F_g= mg$. Let us write the projections of forces on the vertical axis:

$N+F\cos(90^\circ-a) = mg \Rightarrow N = mg - F\sin a.$ The force of friction is $F_{\text{fr}} = \mu N = \mu (mg-F\sin\alpha)$ .

The total acceleration is 0, so $F\cos a - F_{\mathrm{fr}} = 0$ . Therefore, $F = \dfrac{F_{\text{fr}}}{\cos a} = \dfrac{\mu (mg-F\sin\alpha)}{\cos\alpha}.$

So $F = \dfrac{\mu m g}{\cos a + \mu \sin a} = 176.8\,\mathrm{N}.$