(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A flagpole of mass 25kg and 2.5m in length is held in a horizontal position by a cable shown in the diagram.

The centre of gravity of the flagpole is at a distance of 1.5m from the fixed end. Determine:

athe tensionTin the cable;

bthe vertical component of the force at the fixed end of the pole.

2. Relevant equations

moment = Fx

x = d sinθ

ac/w moment = c/w moment

F_{y}= F sinθ

3. The attempt at a solution

I attempted partaand think I have got the answer, however I would be grateful for somebody to verify this for me.

If we call the distance between the fixed end of the flagpole to its centre of gravity x_{1}and its weight F_{1}then moment_{1}= F_{1}x_{1}= 25 x 9.81 x 1.5 = 367.875 Nm.

If we then call the length of the flagpole d and the tensionTF_{2}, and we know that x_{2}= d sinθ, then moment_{2}= F_{2}x 2.5 sin30 = F_{2}x 1.25 and therefore F_{2}= moment_{2}/ 1.25.

The principle of moments states that when an object is in equilibrium , the anticlockwise moments equal the clockwise moments and therefore moment_{1}= moment_{2}.

Therefore the tensionT= 367.875 / 1.25 = 294.3 N.

Where I am really stumped is partb. I know how to work out the vertical component of a force - F_{y}= F sinθ - that's pretty easy. However when the question asks forthe force at the fixed end of the polei'm not quite sure what that is and thus I can't work out the vertical component.

If anybody could enlighten me of what that force is I'd be very pleased.

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# Homework Help: Physics Homework: Moments and Tension

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