Moment Physics

Suppose your feet were glued to the floor, and someone tries to knock you over. Would they try and push at your ankles or at your shoulders? Assuming you don't want to fall over, you would want him to push at your ankles because this way he can exert only a small moment on you because of the small distance to the pivot point at your feet, and it is not the force but it is the moment he exerts that will make you turn around your pivot (your feet) and fall.

Let's go back to the figure above. If we push in the indicated direction at a distance of$5\mathrm{m}$from the pivot, then the perpendicular distance will be roughly$4\mathrm{m}$. If we push with a force of$100\mathrm{N}$at this distance in this direction, then we exert a moment of$400\mathrm{Nm}$.

Suppose somebody is stuck in an elevator and you need to break down the door to rescue them. The force at which the door breaks is$4000\mathrm{N}$. This is a lot more than you can exert with your muscles, so you get a crowbar which gives you leverage. If the crowbar is as is depicted in the illustration below, how much force do you need to exert on the crowbar in order to break the door?

A crowbar (green) is used to break a door (to the right) by using a wall (to the left) to stabilise its pivot (red dot), and where you exert the force F, StudySmarter Originals.

Well, we see that we need to exert a moment of$4000\mathrm{N}\times 5\mathrm{cm}=200\mathrm{Nm}$on the door, so the force we need to exert on the crowbar is

$F=\frac{M}{d}=\frac{200\mathrm{Nm}}{1\mathrm{m}}=200\mathrm{N}$.

Suddenly, this force is very realistic for a person to exert on an object, and we are able to break the door.

Alice and her dad Bob are sitting on a seesaw and want to make it balance. Alice is lazy and does not want to move, so she stays a distance of$2\mathrm{m}$away from the pivot. The mass of Alice is$20\mathrm{kg}$and the mass of Bob is$80\mathrm{kg}$. At what distance from the pivot does Bob need to sit in order for the seesaw to be balanced?

Answer: For a balanced seesaw, the moments on the seesaw have to cancel each other out, so$M_{\mathrm{Alice}}=M_{\mathrm{Bob}}$. The force on the seesaw is perpendicular to the horizontally balanced seesaw, so the perpendicular distance$d$is equal to the distance of the person to the pivot. This means that for a balanced seesaw, we require

$m_{\mathrm{Alice}}g{d}_{\mathrm{Alice}}=m_{\mathrm{Bob}}g{d}_{\mathrm{Bob}}$.

The factor of the gravitational field strength cancels out (so this problem has the same answer on other planets too!), and we calculate

$d_{\mathrm{Bob}}=\frac{m_{\mathrm{Alice}}{d}_{\mathrm{Alice}}}{m_{\mathrm{Bob}}}=\frac{20\mathrm{kg}\times 2\mathrm{m}}{80\mathrm{kg}}=0.5\mathrm{m}$.

We conclude that Bob needs to sit a distance of$0.5\mathrm{m}$away from the pivot. This makes sense: Alice needs 4 times as much leverage as Bob to compensate for her weight being 4 times as small as Bob's weight.

Suppose you have a spanner and you want to know the size of the moment it takes to undo a certain nut. You get a machine that delivers a constant large force, say$1000\mathrm{N}$, and a string such that you can exert a force on the spanner at a very specific place. See the illustration below for the setup. You then start by placing the string as close to the nut (the middle of which is the pivot) as possible. Chances are the spanner doesn't move, because the distance$d$is so small that the moment on the spanner is also small. Slowly you move the string further and further away from the nut, thereby exerting a larger and larger moment on the nut through an increasing perpendicular distance of the force to the pivot. At some distance$d$to the pivot, the nut starts to turn. You record this distance$d$to be$6\mathrm{cm}$. Then the moment you exerted on the nut was$M=1000\mathrm{N}\times 6\mathrm{cm}=60\mathrm{Nm}$. You conclude that it takes a moment of about$60\mathrm{Nm}$to undo this particular nut.

A spanner and a nut, with the pivot, string, and the force delivering machine, StudySmarter Originals.