Question

2. the uniform 150-lb beam is initially at rest when the forces are applied to the cables. determine the magnitude of the acceleration of the mass center and the angular acceleration of the beam at this instant.

Answer 1

The acceleration of the mass center is $a = 7.857 rad / s^2$

and the angular acceleration of the beam at this instant, $a_{G} = 34.1 ft/ s^2$

The mass moment of inertia of the beam :

Thus Ig  = 1/2 ml^2

this implies Ig = 1/2 ( 150/ 32.2 ) (12^2)

Thus the required value of Ig = 55.90 slug ft^2

Sigma f of x = m(a g)x

which is equal to Sigma f of x  = 21.47 ft/s^2 --------- 1

because 200 cos 60 = 150/ 32.2 (ag)x

and Sigma f of y = m a (g)y = 26.45 ft^2 --------2

because 200 sin 60 = 55.90α + 100(6)

The acceleration of the mass centre is a = 7.857 rad / s^2

Thus the magnitude of the acceleration is

aG = $\sqrt{21.47^2+ 26.45^2}$

aG = 34.1 ft/ s^2

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