Efficiency of inclined plane

EFFICIENCY OF INCLINED PLANE

• The efficiency of an inclined plane is defined as the ratio between the effort without friction (P0) and the effort with friction (P).

• Mathematically,

1. Efficiency for the Motion Up the Plane

Dividing equation (6.6) by equation (6.8), we get

Multiplying the numerator and denominator by sin (α + ϕ) sin θ, we get

Note

1. When effort is applied horizontally, then θ = 90°.

2. When effort is applied parallel to the plane, then θ = 90° + α.

2. Efficiency for the Motion Down the Plane

Since the value of P will be less than P0, dividing equation (6.10) by equation (6.9), we get

Multiplying the numerator and denominator by sin (α – ϕ) sin θ, we get

Note

1. When effort is applied horizontally, then θ = 90°.

2. When effort is applied parallel to the plane, then θ = 90° + α.

Example 6.2

 An effort of 1200 N is required just to move a inclined plane of angle 12, the force acting parallel to the plane. If the angle of inclination of the plane is increased to 15°, then the effort required is 1400 N. Find the weight of the body and coefficient of friction.

Given data: 

P1 = 1200 N; α1 = 12°; P2 = 1400; α2 = 15°.

Solution: In both the cases, the effort is applied parallel to the inclined plane and body is just to move up. Hence the force of friction F = μ R will be acting downwards.

Coefficient of friction (μ):

Case (i): The body is in equilibrium under the action of forces, as shown in Fig.6.9(a). 

P1 = 1200 N; ά1 = 12° (given)

Let F1 and RNI be the force of friction and normal reaction respectively. 

Resolving the forces along the plane, we get

W sin 12° = F1 = 1200

or W sin 12° + μ RNI = 1200

[ ⸪ F1 = μ RN1]  ... (i)

Resolving the forces normal to the plane, we get

RNI = W cos 12°

Substituting the value of RN1 in equation (i), we get

Case (ii): The body is in equilibrium under the action of forces, as shown in Fig.6.9(b). 

P2 = 1400 N and α2 = 15° (given)

Let F2 and RN2 be the force of friction and normal reaction respectively.

Resolving forces along the plane [refer Fig.6.9(b)], we get

Resolving forces normal to the plane, we get

RN2 = W cos 15°

Substituting the value of RN2 in equation (iii), we get

Weight of the body (W):

The weight of the body is obtained by substituting the value of u in equation (ii) (or in (iv)).