A trolley is falling freely on inclined plane | vedclass

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Question

(c)

Component of tension force along the plane $= T \cos \alpha$

Component of weight along the plane $= mg \sin 	heta$

Acceleration of the c

Vedclass · Accepted Answer

$90^{\circ}$

Vedclass · Answer

(c)

Component of tension force along the plane $= T \cos \alpha$

Component of weight along the plane $= mg \sin 	heta$

Acceleration of the car = Acceleration of the bob in the car as the bob is in equilibrium in frame of the car

So, net force along the plane

$m g \sin 	heta+T \cos \alpha=m a$

$mg \sin 	heta+ T \cos \alpha= m (g \sin 	heta)$

$T \cos \alpha=0$

But $T$ is not equal to zero so,

$\cos \alpha=0$

Then,

$\alpha=90^{\circ}$

A trolley is falling freely on inclined plane as shown in the figure. The angle of string of pendulum with the ceiling of trolley is $(\alpha)$ equal to ..........

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