Question

a uniform rod 120 cm long is at its center. A load of 800 gm is hung from its left end. A force is applied at a distance of 48 cm from the end of the rod. if the rod remains at the horizontal position what is magnitude of this force

Answer 1

Answer:

At horizontal position the reaction 'N'.

Let the mass of rod is m.

The moment of inertia of rod about the axis passig through the end and perpendicular to its plane is

I=31ml2

For rotation of rod,

Moment of force = Applied torque

⟹x×F=Iα

⟹α=31ml2xF=ml23xF

Here α is angular acceleration about hinge .

The position of centre of mass is aat the middle.

So the linear acceleration of centre of mass,a=α2l

a=2ml3xF

So, the net force in vertical on centre of mass is

=ma

=2l3xF

This should be equal to N.

So, N=2l3xF

Answer 2

Answer:

ANSWER

At horizontal position the reaction 'N'.

Let the mass of rod is m.

The moment of inertia of rod about the axis passig through the end and perpendicular to its plane is

I=31ml2

For rotation of rod,

Moment of force = Applied torque

⟹x×F=Iα

⟹α=31ml2xF=ml23xF

Here α is angular acceleration about hinge .

The position of centre of mass is aat the middle.

So the linear acceleration of centre of mass,a=α2l

a=2ml3xF

So, the net force in vertical on centre of mass is

=ma

=2l3xF

This should be equal to N.

So, N=2l3xF

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