Question

A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force $F_1$ and the other hand holding it up at 50 cm from the end of the plank with force $F_2$. If the plank has a mass of 20 kg and its center of gravity is at the middle of the plank, what are the magnitudes of the forces $F_1$ and $F_2$ ?

$F_1 = 196 \textrm{ N}$, $F_2 = 392 \textrm{ N}$

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This is College Physics Answers with Shaun Dychko. We have a two meter long plank of wood and we’re told that the center of mass is exactly in the center, so that’ll be at a position two meters divided by two, which is 1.0 meters from an end. That’s also going to be the level arm of this gravity force when we calculate its torque above the pivot here at the end. We don’t have to choose the pivot to be there but it is convenient to do so because when we have our torque equation, it’s going to consist only of force

*F2*and*Fg*and it will not have force*F1*in it because the level arm for force one is zero if we choose this to be the pivot point. So, what else do we know. We know that*r2*is 0.5 meters, that’s the level arm for force two, and the mass of the plank is 20 kilograms. So the first condition for static equilibrium is that the total net forces have to be zero. So the total forces up have to equal the total forces down. And there's only one force upwards, that’s*F2*, and there are two forces downwards,*F1*plus*Fg*, and*Fg*is*mg*. That’s as far as we can go there because there are two unknowns and so we can’t do anything more with that equation,*F2*and*F1*are both things we don’t know. Now let’s consider torque. The total counter-clockwise torque has to equal the total clockwise torque. And*F2*is exerting a torque that is counter-clockwise, so we have*F2*multiplied by*r2*on the left side, on the right side we have*mg*times*rg*is the torque exerted by gravity, and this has only one unknown there which is*F2*so we’ll solve for it by dividing both sides by*r2*. And so we have force two is*mg*, level arm of gravity divided by level arm of force two, so that’s 20 kilograms times 9.8 newtons per kilogram times one meter divided by 0.5 meters which gives 392 newtons. Then we can rearrange this formula where we talked about net force and solve it for*F1*by subtracting*mg*from both sides and we get that*F1*is*F2*minus*mg*, so that’s 392 newtons that we just calculated minus 20 kilograms times 9.8, giving us 196 newtons.