physics question

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rinsterman12

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A crane has a mass of 4000kg and a base of 3.4 m. The arm of the crane is 22m and attaches to the center of the crane. IF the arm is placed at an angle of 30 degrees, what is the largest mass that the crane can hold off the ground without tipping? I keep getting 4000 kg. I place my fulcrum at the center of the crane arm. Can someone help me with this problem. The only thing I am having trouble with is placing the pivot point? why can it not be at the center of the crane arm?

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physics is currently owning me so i don't know if this is right...

your answer would be ok if there were indeed a fulcrum halfway up the crane arm. but there's not. loading the end of the crane changes where the normal force for the whole crane has to be located to oppose the whole rig. when the normal force actually is located out from under the crane base, it tips. the edge of the crane base towards the arm and hanging weight is the fulcrum around which the crane will rotate when it tips.

so i'd be tempted to center my calcs there and do 4000*basemgtobaseedge and X*hangingweighttobaseedge and set them equal.

anyone else confirm or deny this train of thought?
 
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Is this a question that requires knowledge f levers? If so then idont know that it's needed. Nowhere in the amcas official outline fr physics are levers mentioned. I'm not sure it's necessary...
 
Because the crane will not pivot at the center. Imagine it has a little too much weight. It will tip on it's edge. The COM acts at the center of the crane, but the point of rotation is on the edge, therefore the lever arm for that side of the fulcrum will be 1.7m. The lever arm for the opposite side will be 22 x cos 30.
 
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