Question

A body resting on a rough horizontal plane requires a pull of 200 N inclined at 30o to the plane just to move it. It was found that a push of 255 N inclined at 30o to the plane just moved the body. Determine the weight of the body and coefficient of friction

Answer 1

Answer:

Given : \theta=30^{\circ}θ=30

∘

Let W = Weight of the body in newtons,

R_{ N }R

N

= Normal reaction,

\muμ = 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,

F=180 \cos 30^{\circ}=180 \times 0.866=156 NF=180cos30

∘

=180×0.866=156N

Now resolving the forces vertically,

R_{ N }=W-180 \sin 30^{\circ}=W-180 \times 0.5=(W-90) NR

N

=W−180sin30

∘

=W−180×0.5=(W−90)N

We know that F=\mu \cdot R_{ N } \quad \text { or } \quad 156=\mu(W-90)F=μ⋅R

N

or 156=μ(W−90) …(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,

F=220 \cos 30^{\circ}=220 \times 0.866=190.5 NF=220cos30

∘

=220×0.866=190.5N

Now resolving the forces vertically,

R_{ N }=W+220 \sin 30^{\circ}=W+220 \times 0.5=(W+110) NR

N

=W+220sin30

∘

=W+220×0.5=(W+110)N

We know that F=\mu \cdot R_{ N } \quad \text { or } \quad 190.5=\mu(W+110)F=μ⋅R

N

or 190.5=μ(W+110) …(ii)

From equations (i) and (ii),

W=1000 N , \text { and } \mu=0.1714W=1000N, and μ=0.1714

Explanation:

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