Equilibrium of a body on a horizontal plane

EQUILIBRIUM OF A BODY ON A HORIZONTAL PLANE

Suppose a body A of weight (W) is resting on a horizontal plane B, as shown in Fig.6.4. Let a tractive force P is applied at an angle θ with the horizontal, the body A just tends to move.

The active forces on the body A are:

(i) Weight, W, and

(ii) Tractive force or effort, P.

The reactive forces are:

(i) Normal reaction, RN, and

(ii) Frictional force, F.

Thus, external forces acting on the body A are:

Vertical:

(i) (W − P sin θ) ↓ (downwards)

(ji) RN ↑ (upwards)

Horizontal:

(i) P cos α → (towards right)

(ii) F = μ • RN← (towards left)

For satisfying the equilibrium conditions, ΣFV = 0 and Σ FH = 0

or P cos θ cos ϕ = W sin ϕ - P sin θ • sin ϕ

or P (cos θ • cos ϕ + sin θ sin ϕ) = W sin ϕ

The value of P is minimum when cos (θ - ϕ) is maximum, so

Hence, the effort P will be minimum if its angle of inclination & with the horizontal is equal to the angle of friction ϕ.

Example 6.1 

The force required to pull a body of weight 50 N on a rough horizontal plane is 15 N. Determine the coefficient of friction, if the force is applied at an angle of 15° with the horizontal.

Given data: W = 50 N; P = 15N, θ = 15°.

Solution: The body is in equilibrium under the action of the forces, as shown in Fig.6.5. Resolving the forces horizontally, we get

F = μ RN = 15 cos 15° ... (i)

Now, resolving forces vertically, we get

RN + 15 sin 15° = W = 50

⸫ RN = 50 - 15 sin 15° = 46.12 N

Substituting the value of RN in equation (i), we get

μ × 46.12 = 15 cos 15° or μ = 0.314 Ans.