Engineering Mechanics

Chapter 8 : Principles of Friction „„„„„ 135

and now resolving the forces along the plane,

200 = F 1 + W sin 15° = μ.R 1 + W sin 15 ° ...(Q F = μ.R )
= μ W cos 15° + W sin 15° ...(Q R 1 = W cos 15°)
= W (μ cos 15° + sin 15°) ...(ii)
Now consider the body lying on a plane inclined at an angle of 20° with the horizontal and
subjected to an effort of 230 N shown in Fig. 8.12 (b).

Resolving the forces at right angles to the plane,

R 2 = W cos 20° ...(iii)

and now resolving the forces along the plane,

230 = F 2 + W sin 20° = μ R 2 + W sin 20° ...(Q F = μ.R)

= μ W cos 20° + W sin 20° ...(Q R 2 = W cos 20°)

= W (μ cos 20° + sin 20°) ...(iv)

Coefficient of friction

Dividing equation (iv) by (ii),

230 ( cos 20 sin 20 )

200 ( cos15 sin 20 )

W

W

μ °+ °

=

μ °+ °

230 μ cos 15° + 230 sin 15° = 200 μ cos 20° + 200 sin 20°

230 μ cos 15° – 200 μ cos 20° = 200 sin 20° – 230 sin 15°

μ (230 cos 15° – 200 cos 20°) = 200 sin 20° – 230 sin 15°

∴

200 sin 20 – 230 sin 15

230 cos 15 – 200 cos 20

°°

μ=

°°

(200 0.3420) – (230 0.2588)

0.259

(230 0.9659) – (200 0.9397)

××

==

××

Ans.

Weight of the body

Substituting the value of μ in equation (ii),

200 = W (0.259 cos 15° + sin 15°)

= W (0.259 × 0.9659 + 0.2588) = 0.509 W

∴

200

392.9 N

0.509

W==^ Ans.

EXERCISE 8.1

1. Find the horizontal force required to drag a body of weight 100 N along a horizontal
   plane. If the plane, when gradually raised up to 15°, the body will begin to slide.
   [Ans. 26.79 N]
   Hint. φ = 15° or μ = tan φ = tan 15° = 0.2679


2. A body of weight 50 N is hauled along a rough horizontal plane by a pull of 18 N acting
   at an angle of 14° with the horizontal. Find the coefficient of friction. [Ans. 0.383]


3. A man is walking over a dome of 10 m radius. How far he can descend from the top of the
   dome without slipping? Take coefficient of friction between the surface of the dome and
   shoes of the man as 0.6. [Ans. 1.413 m]