Calculate how much weight the crane could lift

AI Thread Summary
To calculate the weight a crane can lift, it's essential to consider the forces acting on the jib, including the mass and acceleration. The jib's different lengths create varying loads, particularly with a 16T counterweight on the shorter end. The mechanics of equilibrium must be analyzed to understand how these forces balance out. A sketch of the crane setup can aid in visualizing the problem and clarifying the calculations. Further assistance in mechanics is needed to resolve the confusion regarding the jib's dimensions and load distribution.
Clueless87
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Homework Statement
Hi all, new to the forum so not sure if I’m filling this out correctly but I’m currently studying for an exam for my work and going through past papers. The question is;

The horizontal jib of a tower crane is 36m long, weights 3T and is mounted on a turntable 8m from the machinery end of the jib. The shorter end of the jib carries a 16T counter weight at its furthest extent.

i)Draw a sketch of the crane (east enough), and the loads exerted on it (not so easy…for me at least) (5 points)

ii) Calculate how much weight the crane could lift if the load were suspended at the far end of the crane’s jib and the jib is in equilibrium. (5 points)
Relevant Equations
I have no idea….
I think the first part is F = mxa? But given the jib is resting on a turntable at a certain distance, does that mean the longer and shorted ends experience different loads? Especially when the shorter end has a 16T counterweight?

The second part, I have no idea how to calculate this nor the equilibrium.

Any help would be greatly appreciated!
 
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Start by making a sketch and post it here!
 
Clueless87 said:
The horizontal jib of a tower crane is 36m long, weights 3T and is mounted on a turntable 8m from the machinery end of the jib. The shorter end of the jib carries a 16T counter weight at its furthest extent.
I can't make sense of that. 8<36/2, yet the other end is shorter? Should it be 28?
 

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