Engineering Mechanics

Chapter 8 : Principles of Friction „„„„„ 141

Substituting the value of R 1 in equation (i),

P 1 cos θ = W sin α – μ (W cos α – P 1 sin θ)

= W sin α – μ W cos α + μ P 1 sin θ

P 1 cos θ – μ P 1 sin θ = W sin α – μ W cos α

P 1 (cos θ – μ sin θ) = W (sin α – μ cos α)

∴ 1

(sin – cos )

(cos – sin )

PW
α μ α
=×
θ μ θ
and now substituting the value of μ = tan φ in the above equation,

1

(sin – tan cos )

(cos – tan sin )

PW

α φ α

=×

θ φ θ

Multiplying the numerator and denominator by cos φ,

1

(sin cos – sin cos ) sin ( – )

(cos cos – sin sin ) cos ( )

PW W

α φφααφ

=× =×

θ φφθθ+φ

2. Maximum force (P 2 ) which will keep the body in equilibrium, when it is at the point of
   sliding upwards.

In this case, the force of friction (F 2 = μ R 2 ) will act downwards as the body is at the point of
sliding upwards as shown in Fig. 8.17 (b). Now resolving the forces along the plane.

P 2 cos θ = W sin α + μ R 2 ...(iii)

and now resolving the forces perpendicular to the plane,

R 2 = W cos α – P 2 sin θ ...(iv)

Substituting the value of R 2 in equation (iii),

P 2 cos θ = W sin α + μ (W cos α – P 2 sin θ)

= W sin α + μ W cos α – μ P 2 sin θ

P 2 cos θ + μ P 2 sin θ = W sin α + μ W cos α

P 2 (cos θ + μ sin θ) = W (sin α + μ cos α)

∴ 2

(sin cos )

(cos sin )

PW

α+μ α

=×

θ+μ θ

and now substituting the vaue of μ = tan φ in the above equation,

2

(sin tan cos )

(cos tan sin )

PW

α+ φ α

=×

θ+ φ θ

Multiplying the numerator and denominator by cos φ,

2

(sin cos sin cos ) sin ( )

(cos cos sin sin ) cos ( – )

PW W

α φ+ φ αα+φ
=× =×
θ φ+ φ θθφ
Example 8.11. Find the force required to move a load of 300 N up a rough plane, the force
being applied parallel to the plane. The inclination of the plane is such that when the same load is
kept on a perfectly smooth plane inclined at the same angle, a force of 60 N applied at an inclination
of 30° to the plane, keeps the same load in equilibrium.

Assume coefficient of friction between the rough plane and the load to be equal to 0.3.