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I was wondering if anyone knew the formula to calculate how much force is required to lift a tire or any object by the edge.

For example if a tire weighs 500 lbs (or has a mass of 226.8 kg). I would just like to know the formula how to calculate how much of that 500 lbs we are actually lifting since it is only from the edge.

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    $\begingroup$ If you lift slowly then it is half. $\endgroup$ – ja72 Mar 4 '14 at 18:17
  • $\begingroup$ Do you mean lifting with part of the object remaining on the ground? Or actually lifting off the ground? $\endgroup$ – Kyle Oman Mar 4 '14 at 18:20
  • $\begingroup$ Related? physics.stackexchange.com/q/52701 $\endgroup$ – user10851 Mar 4 '14 at 20:11
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If you are lifting on one edge and it is resting on the other edge, and the edges are an equal distance from the center of mass, then the answer is $$\boxed{F = \frac{1}{2} W}$$

If you are lifting with a distance of $\ell_1$ from the center, and the pivot is $\ell_2$ from the center then $$\boxed{ F = \frac{\ell_2}{\ell_1+\ell_2} W } $$

This is commonly known as the lever rule.

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It will in general depend on the shape of the object. If it has a large concentration of mass at the edge you are lifting then the force will be close to its weight; if its mass is concentrated near the other edge then it will be very small.

The general case is solved using the law of levers:

enter image description here

If $d$ is the distance from the fulcrum to the centre of mass and $L$ is the distance to the edge you're lifting, then the force $F$ that you apply is related through the weight $W$ of the object through $$ LF=dW $$ at equilibrium. For most symmetric objects, like tyres, the centre of mass will be at the geometric centre of the object, and therefore you will need to apply half the object's weight to hold it, and slightly more than that to lift it.

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    $\begingroup$ That's why you put the bags of concrete mix near the wheel of the wheelbarrow, not near the handles... $\endgroup$ – DJohnM Mar 4 '14 at 18:36

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