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 αNow substituting the value of μ = tan φ in the above equation,
1(sin – tan cos )
(cos tan sin )PWα φ α
=×
α+ φ α
Multiplying the numerator and denominator by cos φ,1sin 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,2sin tan cos
cos – tan sinPW
α+ φ α
=×
α φ α
Multiplying the numerator and denominator by cos φ,2sin cos sin cos
cos cos – sin sinPWα φ+ φ α
=×
α φφαsin ( )
cos ( )Wα+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.