Chapter 8 : Principles of Friction 137

`P 1 (cos α + μ sin α) = W (sin α – μ cos α)`

`∴ 1`

`(sin – cos )`

(cos sin )

`PW`

α μ α

=×

α+μ α

Now substituting the value of μ = tan φ in the above equation,

`1`

`(sin – tan cos )`

(cos tan sin )

`PW`

`α φ α`

=×

α+ φ α

Multiplying the numerator and denominator by cos φ,

`1`

`sin cos – sin cos sin ( – )`

cos cos sin sin cos ( – )

`PW W`

α φφααφ

=× =×

α φ+αφ α φ

= W tan (α – φ) ...(when α > φ)

= W tan (φ – α) ...(when φ > α)

- 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 = μR 2 ) will act downwards, as the body is at the point of

sliding upwards as shown in Fig.8.12. (b). Now resolving the forces along the plane,

`P 2 cos α = W sin α + μR 2 ...(iii)`

and now resolving the forces perpendicular to the plane,

R 2 = W cos α + P 2 sin α ...(iv)

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

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

= W sin α + μ W cos α + μ P 2 sin α

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

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

`∴ 2`

`(sin cos )`

(cos – sin )

`PW`

α+μ α

=×

α μ α

Now substituting the value of μ = tan φ in the above equation,

`2`

`sin tan cos`

cos – tan sin

`PW`

α+ φ α

=×

α φ α

Multiplying the numerator and denominator by cos φ,

`2`

`sin cos sin cos`

cos cos – sin sin

`PW`

`α φ+ φ α`

=×

α φφα

`sin ( )`

cos ( )

`W`

`α+φ`

=×

α+φ

= W tan (α + φ)

Example 8.8. An object of weight 100 N is kept in position on a plane inclined 30° to the

horizontal by a horizontally applied force (F). If the coefficient of friction of the surface of the

inclined plane is 0.25, determine the minimum magnitude of the force (F).

Solution. Given: Weight of the object (W ) = 100 N; Angle at which plane is inclined

(α) = 30° and coefficient of friction (μ) = 0.25 = tan φ or φ = 14°.

We know that the minimum magnitude of the force to kept the object in position (when it is at

the point of sliding downwards),

`F = W tan (α – φ) = 100 tan (30° – 14°) = 100 tan 16°`

= 100 × 0.2867 = 28.67 N Ans.