Engineering Mechanics

Chapter 9 : Applications of Friction „„„„„ 155

Dividing both sides by 3 cos α,

1000 + 100 = Fw + Rw tan α

1100 = 99.1 + 385.4 tan α

385.4 tan α = 1100 – 99.1 = 1000.9

∴ tan 1000.9 2.5970

385.4

α= = or α = 68.9° Ans.

Example 9.5. Two identical blocks of weight W are supported by a rod inclined at 45° with
the horizontal as shown in Fig. 9.7.

Fig. 9.7.
If both the blocks are in limiting equilibrium, find the coefficient of friction (μ), assuming it
to be the same at floor as well as at wall.

Solution. Given: Weight of blocks A and B = W and inclination of rod with the horizontal
(α) = 45°.

Let μ = Coefficient of friction, and

l = Length of the rod.

The forces acting on both the blocks are shown in Fig. 9.8

Resolving the forces vertically.

Fw + Rf = 2W

or μRw + Rf = 2W ...(Q Fw = μRw) ...(i)

and now resolving the forces horizontally.

Rw = Ff = μRf ...(ii)

Now substituting this value of Rw in equation (i),

μ (μRf) + Rf = 2W

μ^2 Rf + Rf = 2W

Rf (μ^2 + 1) = 2W

∴^22

1

f

W

R =

μ +

...(iii)

and now substituting this value of Rf in equation (ii),

2

2

1

w

W

R =μ×

μ + ...(iv)