Engineering Mechanics

(Joyce) #1

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]

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