Mechanics question

**Big_Joe** · May 21st 2009, 07:19 AM

Hey i am stuck on this question i have found in a past paper can someone please help me solve this asap as i have a mechanics exam tomorrow and i need to be able to do all topics as well as possible:

A steel girder AB has a weight 210N. It is held in equilibrium in a horizontal position by two vertical cables. One cable is attatched to the end A. The other cable is attatched to the point C on the girder, where AC = 90cm. The girder is modeled as a uniform rod, and the cables as light inextensible strings.

Given that the tension in the cable at C is twice the tension in the cable at A, show that AB = 120cm.

Thanks to anyone who helps.

**e^(i*pi)** · May 21st 2009, 08:23 AM

Originally Posted by Big_Joe

Hey i am stuck on this question i have found in a past paper can someone please help me solve this asap as i have a mechanics exam tomorrow and i need to be able to do all topics as well as possible:

A steel girder AB has a weight 210N. It is held in equilibrium in a horizontal position by two vertical cables. One cable is attatched to the end A. The other cable is attatched to the point C on the girder, where AC = 90cm. The girder is modeled as a uniform rod, and the cables as light inextensible strings.

Given that the tension in the cable at C is twice the tension in the cable at A, show that AB = 120cm.

Thanks to anyone who helps.

Did you draw a diagram?

From mine you can see where A, B and C with relation to T1 and T2. From the question you can deduce $T_2 = 2T_1$

Resolve vertically about point C and let from C to A be the positive direction:

$mg - x = 2 \times 90 \rightarrow 210-x=180$

$x = 30$

From the diagram we can see that AB = AC + CB = 90+30 = 120cm

Edit1: oops my diagram is wrong - I'll upload the correct one soon
Edit 2: done

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