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## Homework Statement

A crane with jib AB, 7m long. Centre of gravity 3m from A, inclined at 30 degrees to the vertical.

The horizontal tie is attached at C, 5m from A.

If the mass of the jib is 200kg and the load at B is 600kg, what is the tension in the tie.

http://img139.imageshack.us/img139/2818/cranezu9.jpg [Broken]

## Homework Equations

[tex]

\sum {M_{CW} } = \sum {M_{ACW} }

[/tex]

## The Attempt at a Solution

OK, well if we take the pivot about the point A, we have;

[tex]

\begin{array}{l}

\sum {M_{CW} } = \sum {M_{ACW} } \\

\left( 3 \right)\left( {200 \times 9.8} \right) + \left( 7 \right)\left( {600 \times 9.8} \right) = \left( 5 \right)\left( {T\sin 30} \right) \\

\Rightarrow T = \frac{{5880 + 41160}}{{2.5}} = 18816N \\

\end{array}

[/tex]

This tension seems very high for the system.

Could someone please explain the principle behind the Tension and angle components of the crane, because I don't think I fully understand it.

Thanks in advance

Steven.

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