A 2m irregular plank weighing W kg is suspended in the manner shown below,  by strings of negligible weight. If the strings make an angle of 35°  and 55° respectively with the vertical, find the location of center of gravity of the plank from the left end.

Asked by Pragya Singh | 1 year ago |  101

Solution :-

The free body diagram of the above figure is:

Given,

Length of the plank, l = 2 m

θ1 = 350 and θ=550

Let  T1 and T2 be the tensions produced in the left and right strings respectively.

So at translational equilibrium we have;

T1sinθ1  = T2sinθ2

$$\dfrac{T_1}{T_2}=\dfrac{sinθ_2}{sinθ_1}=\dfrac{ sin55}{ sin35}$$

$$\dfrac{T_1}{T_2}=\dfrac{0.819}{0.573 }$$= 1.42

T1 = 1.42T2

Let ‘d’ be the distance of the center of gravity of the plank from the left.

For rotational equilibrium about the centre of gravity :

T1cos35 x d = Tcos 55 (2 – d)

($$\dfrac{T_1}{T_2}$$)x0.82d = (2 x 0.57 – 0.57d)

Substituting T1 = 1.42T2

1.42 x 0.82 d + 0.57 d = 1.14

1.73d = 1.14

Therefore d = 0.65m

Answered by Abhisek | 1 year ago

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