A body, resting on a rough horizontal plane required a pull of 180 N inclined at 30º to the plane just to move it. It was found that a push of 220 N inclined at 30º to the plane just moved the body. Determine the weight of the body and the coefficient of friction.

Question

A body, resting on a rough horizontal plane required a pull of 180 N inclined at 30º to the plane just to move it. It was found that a push of 220 N inclined at [latex]30^{\circ}[/latex] to the plane just moved the body. Determine the weight of the body and the coefficient of friction.

Accepted Answer

Given : [latex] heta=30^{\circ}[/latex]

Let W = Weight of the body in newtons,

[latex]R_{ N }[/latex] = Normal reaction,

[latex]\mu [/latex] = Coefficient of friction, and

F = Force of friction.

First of all, let us consider a pull of 180 N. The force of friction (F) acts towards left as shown in Fig. (a).

Resolving the forces horizontally,

[latex]F=180 \cos 30^{\circ}=180 imes 0.866=156 N[/latex]

Now resolving the forces vertically,

[latex]R_{ N }=W-180 \sin 30^{\circ}=W-180 imes 0.5=(W-90) N[/latex]

We know that [latex]F=\mu \cdot R_{ N } \quad ext { or } \quad 156=\mu(W-90)[/latex] ...(i)

Now let us consider a push of 220 N. The force of friction (F) acts towards right as shown in Fig.(b).

Resolving the forces horizontally,

[latex]F=220 \cos 30^{\circ}=220 imes 0.866=190.5 N[/latex]

Now resolving the forces vertically,

[latex]R_{ N }=W+220 \sin 30^{\circ}=W+220 imes 0.5=(W+110) N[/latex]

We know that [latex]F=\mu \cdot R_{ N } \quad ext { or } \quad 190.5=\mu(W+110)[/latex] ...(ii)

From equations (i) and (ii),

[latex]W=1000 N , ext { and } \mu=0.1714[/latex]

Question : A body, resting on a rough horizontal plane required a pull ...