Question

a uniform rod has a weight 40N and a length 1 m.​

Answer 1

Dear student ,

Here in this case the sum of torque is equal to zero .

So , ∑τ=−r1→×F1−→ + r2→×F2−→=0⇒r1→×F1−→=r2→×F2−→⇒τ2→=0⋅5×40×sin90°=20 NmSo , ∑τ=-r1→×F1→ + r2→×F2→=0⇒r1→×F1→=r2→×F2→⇒τ2→=0·5×40×sin90°=20 Nm

Now look this is the magnitude of torque exerted by the string about a horizontal axis and perpendicular to the torque .

Now the force on the string is the is ,

F=torquelength of the rod=τ2l=20 Nm1 m=20 NF=torquelength of the rod=τ2l=20 Nm1 m=20 N

Now to find the three forces look there are only three forces acting on the rod and this three forces are vertical .

Now the hinge force is , F'=Torquelength of the rod=201=20 NF'=Torquelength of the rod=201=20 N

Regards

ALOK MISHRA{@ALOK16022006}

Answer 2

Answer:

Explanation:

The torque exerted by the string is 20 Nm.

(3) is correct option.

Explanation:

Given that,

Weight of rod= 40 N

Length of rod = 1 m

Rod is at rest. it is in rationally equilibrium

So net torque about point A is zero.

We need to calculate the torque exerted by the string

Using formula of torque

Where,  is the torque exerted by the string

Put the value into the formula

Hence, The torque exerted by the string is 20 Nm.

hope it helps

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