Question

what is the net torque acting on a rod capable of rotating about o as shown in fig​

Answer 1

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Answer 2

Correct Question:

Two forces 50 N and 100 N are acting on a rod capable of rotating it about "O" as shown in Fig. What is the net torque acting on the rod ?

Calculation:

First of all, we need to break down 100 N force into its perpendicular components.

So, Y component of 100 N:

$\therefore \: F_{y} = 100 \sin( {30}^{ \circ} )$

$\implies \: F_{y} = 100 \times \dfrac{1}{2}$

$\implies \: F_{y} =50 \: N$

So, X component of 100 N:

$\therefore \: F_{x} = 100 \cos( {30}^{ \circ} )$

$\implies \: F_{x} = 100 \times \dfrac{ \sqrt{3} }{2}$

$\implies \: F_{x} = 50\sqrt{3} \: N$

Now, from the diagram, we can understand that the X component of the force will not exert any torque because it passes through the axis of rotation.

So, let net torque be $\tau$.

$\therefore \: \tau = \tau_{1} + \tau_{2}$

$\implies\: \tau = (F_{y} \times d_{1})+ (F \times d_{2})$

$\implies\: \tau = (50 \times 0.5)+ (50 \times 0.5)$

$\implies\: \tau = 25 + 25$

$\implies\: \tau = 50 \: Nm$

So, the net torque about point "O" is 50 Nm.