In physics, the teacher showed us a question about balancing torque using an 8-foot seesaw that is 30lb with a 100lb person on one side, asking us to find how far over we would need to move the fulcrum for it to be in equilibrium. Then, when finding the torques, we had the torque with the person being (4-x)(100lb) and the empty side as (x)(30lb) where x is the distance from the fulcrum to the center. I don't understand why we are able to use the radius as x, rather than from the fulcrum to the other side, (x+4). And also why we don't have to compensate for the length of the seesaw (4-x) on the other side of the fulcrum, because I thought that weight would also apply a torque.
2
-
4$\begingroup$ This question would greatly benefit from the addition of a diagram. $\endgroup$– BioPhysicistCommented Feb 17, 2019 at 4:14
-
$\begingroup$ There is an algebraic derivation that proves that calculating torques relative to center of mass gives the correct answer, but that derivation is long. $\endgroup$– David WhiteCommented Jul 30, 2021 at 18:12
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
0
At static equilibrium the net force acting on the system is $0$, and the net torque about any point is also $0$.
I don't understand why we are able to use the radius as x, rather than from the fulcrum to the other side, (x+4).
We could definitely do it this way. But it is more work for us. There are three forces here: the weight of the seesaw, the weight of the person, and the normal upward force supplied by the fulcrum. By choosing the point we calculate the torque about to be the fulcrum, we only have to consider the two weights because the torque of the normal force about the location of the fulcrum is $0$. Then we can say, magnitude wise, the torques due to the weights must be equal.
And also why we don't have to compensate for the length of the seesaw (4-x) on the other side of the fulcrum, because I thought that weight would also apply a torque.
We could definitely do it this way. We could break up the seesaw into two pieces divided by the fulcrum, but once again this would be more work for us than needed. Since we assume the mass of the seesaw is distributed uniformly, we can treat the force of gravity acting on it as if it is acting at the center of the seesaw. Therefore, the mass of the seesaw on the other side of the fulcrum is already considered by considering the entire weight of the seesaw to be a applied at its center.
P.S. As you can tell, working through the majority of statics problems is mostly about efficient calculations. The first sentence of my answer is all you really need to know, so if you are applying that idea then you are never wrong. However, there is always more than one way to tackle a statics problem, and in this case the way the problem was taught to you seems to be the most efficient way to go.
answered Feb 17, 2019 at 4:21
$\begingroup$
$\endgroup$
I'm going to assume that you don't want a calculus explanation because if you were in an calc-based physics, the teacher could do the derivation for you.
Imagine the board is composed of a gazillion ($N$) equal masses ($m$) all linearly positioned, rigidly joined together to form a board of length L and total mass $M$. So now each $m$ has with $L/N$ and mass $m=M/N$, and the position of the middle of each mass is $(j-1/2)\frac{L}{N}$ from one end (let's say the left) of the board, where $j$ designates a specific particle $1\le j \le N$.
Now, if the fulcrum is at position $x$ left of the center, it is $\frac{L}{2}-x$ from the left end. This means that each small mass is at a position, relative to the fulcrum $$p_j=\frac{L}{N}(j-1/2)-L/2+x$$ where a negative value shows it is left of the fulcrum and positive is right of the fulcrum.
If we have a rigid board, the torque produced by each tiny mass is $mgp_j \sin\theta$ where $\theta$ is the angle of the board with respect to vertical. If we add all these torques together we get
$$\Gamma =\frac{M}{N}g\sum_j\left(\frac{L}{N}(j-1/2)-L/2+x \right)$$ $$~~=\frac{M}{N}g\left(\frac{L}{N}\sum_j j-\frac{L}{N}\sum_j 1/2-\sum_jL/2+\sum_jx \right)$$ $$~~=\frac{M}{N}g\left(\frac{L}{N}\frac{N(N+1)}{2}-\frac{L}{N}N/2-\frac{NL}{2}+Nx \right)$$ $$~~=\frac{M}{N}g\left(\frac{L}{2}(N+1)-\frac{L}{2}-\frac{NL}{2}+Nx \right)$$ $$\Gamma=Mgx$$
All the torques from little pieces of mass add up to the same result as the weight acting through the center of mass at a distance $x$ from the fulcrum.
