Work is equal to the force applied to an object times the displacement of the object. (To keep things simple, I’m assuming that the force is applied in the direction of the displacement — no cos thetas — and that the force causes all of the displacement.)
If the object doesn’t move, displacement = 0, and no work was done. So no matter how hard you push on the wall, you’re not doing work. Cool. I get it.
My question is whether displacement is equal to the entire distance the object moves (which makes sense) or to the distance between the object’s starting point and ending point (which is what a number of references I’m finding say).
Because if the latter is true, then pushing a box around in a circle and returning it to its starting point would mean displacement = 0, and no work was done.
I’m not buying that, but I can’t find any sources that say it’s true or that specifically say it isn’t true.
Displacement is a vector and the difference between the final position and the initial position. I should also add, I wouldn’t use “x” to indicate “multiplication” here precisely because the variables involved are vectors and the cross product isn’t correct.
Work done by a force = \int \vec{F} \cdot d\vec{s}
a) Only Component in the direction of motion.
b) If motion is resolved add each comp × disp: F_x\Delta x+F_y\Delta y+F_z\Delta z
c) No work by Force ⟂ to displacement.
Correct. Imagine the circle of movement lies in a vertical plane. You start with a mass at the top of the circle, with gravity pulling directly down on it. As you lower the weight, the mass does work on you. Eventually the mass reaches the bottom of the circle, and now as it continues along the circle back up to the top, you are doing work on the mass. The mass did some work on you, and then you did some work on the mass, and by the time the mass returns to its starting point at the top of the circle, you have done a net total of zero work.
Agreed - it’s just that gravity conveniently defines a force of constant magnitude and direction, regardless of the direction of movement.
This is critical. The friction force vector is always opposite to the instantaneous velocity vector. If you are shoving a box around a circular path on your kitchen floor, then yes, you definitely will have done some work by the time the box returns to its starting point.
Work = Force x Displacement. The box returned to the same position on the kitchen floor, so displacement = 0, so work = 0. How do I factor friction into that?
ETA: To be clear, I believe you. I just don’t get it.
Try to push that box in a circle, or even get it to budge, without ever applying force in the direction of motion. Any non-zero friction will do work on the box whenever it is not standing still. Note that you have to worry about the forces continuously, not just before and after (hence Feynman’s integral sign/differential notation)
Divide the long circular journey around your kitchen floor into a large number of small linear steps.
For each linear step, you move some small distance dx, and the force of friction is directed opposite to your direction of motion. So for each linear step, you will have done F * dx amount of work.
The total path length around the circle is the sum of all those little dx steps. So in the end, the total amount of work you’ve done is F * (sum of all dx), or F * 2 * pi * r.
Going back to my earlier scenario of moving a weight around a vertical circle, the math works out differently than for the case of a sticky kitchen floor. The key is that for each little linear step along the journey around a vertical circular path where there’s no friction (only gravity), the amount of work you do is the component of the force parallel to the direction of movement, multiplied by the distance moved. There are only two spots on this circular journey (at the leftmost and rightmost points of the circle) where the gravitational force vector is exactly parallel to the direction of movement; at those points, the increment of work you do is indeed exactly F * dx. At the top and bottom of the circle, when you advance the weight a distance dx along the circular path, the direction of movement is perpendicular to the direction of the gravitational force vector, so the amount of work you do is zero. If you add up the increments of work you do as you inch your way around a complete vertical circle, the net result is zero work done.
Yeah, I’m trying to explain this to a high-school freshman, so integrals are right out.
I get that logically I have to apply a force to overcome friction. But logically, I did a lot of work to move that box in a circle, except I didn’t because there’s no displacement. So clearly logic doesn’t apply here.
Given that the formula for work is F x d, and that d = 0, how can I calculate the work done to be anything other than 0?
It’s clear what you are saying, but the situation is not the same before and after. The energy did not disappear; instead the friction has heated up some molecules so they are now moving faster. Logic still applies
Well, trying to teach Physics without calculus is a challenge in itself. But, if you understand Trigonometry, it is a bit easier.
Look at the issue with polar-coordinates. Instead of F x d, it becomes F x r x theta, where r is radius and theta is in radians. If you push you box in a full circle, then theta is equal to 2(pi), so W = F x 2(pi)r.
Is the disconnect that work in this question is relevant to the box, not to the person generating the force? In this case, the person’s muscles are doing work since the muscular motion is not related to the path of the box.
In fact, if you want to know what work is done while pushing it in a circle you have to do more than using theta, you have to do an integral. And you will find you have done work and if the box now has zero velocity friction did the exact same work, only it was negative.
One important thing here is to point out that, strictly speaking, work isn’t F x d, at least not with “x” meaning simple scalar multiplication. Work is actually the dot product of the force vector and the displacement vector:
In the end, this boils down to taking the magnitude of the force, the magnitude of the displacement, and doing simple multiplication between the two, but then also multiplying the result by the cosine of the angle between those two vectors:
W = F * dx * cos(theta)
If integrals are out, that’s fine. My previous post takes them through the thought process required to understand what’s happening without mentioning integrals at all; they don’t need to have any calculus training, they only need to understand the idea of breaking the circular movement up into small linear movements (which together in sequence, approximate a circle), calculating the work for each of those linear movements, and then adding up the total.
If you can carry out that process for the scenario of pushing a box around a circular path on your kitchen floor, you’ll see that each increment of movement has you doing an increment of work, so that by the time you return to your starting point, you have done a net non-zero amount of work.
If it helps clear this up in your head, scrap the circular path and just move the box in a straight line across your kitchen floor from point A to point B, and then back to A again. If A and B are separated by distance X, then the total work moving from point A to point B is FX, since you were pushing in the same direction the box moved. When you turned around and pushed the box back to A, you were pushing in the other direction (toward A), so once again force and displacement were pointed in the same direction, and you did another chunk of work, FX. So by moving the box from A to B to A across your kitchen floor, you did 2FX units of work.
Now consider the box sitting on the floor at point A, but now we’ll lift it up to point C, which is a distance X above the floor. You pushed up with force F, the box moved up by distance X, F and X were pointed in the same direction, so you did FX units of work. Now lower the box back down to the floor. Displacement is pointed down, but - and here’s the key - you are still exerting an upward (not downward) force equal to the weight of the box). In this movement, force and displacement are pointing in opposite directions, so you are doing negative work; the amount of work you did for this downward movement is -FX. Add up the two chunks of work, and you did FX - FX = 0 work.
This example is exactly what I was about to post. d is not zero; the box was displaced, work was done.
Just because the object being worked on was returned to the same place, it doesn’t mean displacement is zero.
You could apply this to rotational motion, in which, say, gears carrying forces return to their original position periodically. Was their displacement zero and no work done? Of course not. Else, how could a watch’s mainspring run a watch?
Wish me luck.)