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I have the question:

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And I am asked to find the tension, T, in the rope.

I have drawn a triangle:

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And have tried to use trigonometry to find T. However, the solutions say that T should be 17.8N and I am not sure how this is achieved.

2 Answers 2

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The key is that the bar is attached to the wall at $A$, so it will not slide up or down. So the only thing needed for equilibrium is that the net torque about point $A$ balance to zero.

The torque due to the bar weight is $$ \tau_b = 20 \mbox{ N} \cdot 0.4 \mbox{ m} = 8 \mbox{ N-m} $$

The torque due to the tension of the rope is $$ \tau_r = T_y \cdot 0.7 \mbox{ m} $$ where $T_y$ is the upward component of the force exerted by the rope $$ T_y = T\sin(40^\circ) $$ Solving $$ T\sin(40^\circ) \cdot 0.7 \mbox{ m}= 8 \mbox{ N-m}\\ T = \frac{8}{0.7 \sin(40^\circ)} \frac{\mbox{ N-m}}{\mbox{ m}}\approx 17.78 \mbox{ N} $$

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    Could I just ask where you got 0.4 m from ? (: – 2017-01-19
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You need to balance the torques on the bar around $A$. The weight of the bar acts at the CG, giving a clockwise torque of ???. The rope has to supply a counterclockwise torque to counteract this.

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    I kind of understand. Could you please elaborate ? And is the clockwise torque the same as the anti - clockwise torque ? (: – 2017-01-19
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    The magnitudes are the same but the signs are opposite. The sum is then zero which is the condition for equilibrium. Where are you having trouble computing the torques? – 2017-01-19