Solving Farmer Lever Problem | Min Force Needed

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To lift a 210kg rock using a lever, the farmer needs to exert a minimum force of 233.3N. The lever's handle is positioned 9 times further from the fulcrum than the distance from the fulcrum to the rock. By applying the principle of moments, the equation F times 9x equals the weight of the rock (2100N) multiplied by x can be used to find the required force. This problem falls under the topic of levers and mechanical advantage in physics. Understanding these concepts can help solve similar problems effectively.
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A farmer uses a lever to list a 210kg rock so he can remove it from the field. The distance, l1, from the lever's handle to the fulcrum is 9 times the distance, l2, from the fulcrum to the rock.
Assume: the fulcrum is between the rock and the handle. the weight of the lever is negligible.
What is the minimum force the farmer must exert on the lever in order to lift the rock?

I have no idea how to approach this problem. Can anyone give me some pointers so i can look for stuff in my book or online. what section would this kind of problem fall in.

thanks in advance
 
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Why not google on "using levers with fulcrum" (omit the quotes) ?
 
donjt81 said:
A farmer uses a lever to list a 210kg rock so he can remove it from the field. The distance, l1, from the lever's handle to the fulcrum is 9 times the distance, l2, from the fulcrum to the rock.
Assume: the fulcrum is between the rock and the handle. the weight of the lever is negligible.
What is the minimum force the farmer must exert on the lever in order to lift the rock?
I have no idea how to approach this problem. Can anyone give me some pointers so i can look for stuff in my book or online. what section would this kind of problem fall in.
thanks in advance

It is not hard if you use elementary algebra ...
Let x be the distance from rock to fulcrum, then 9x is from handle to fulcrum

Hence, F times 9x = 2100x

F = 233.3N
 

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