Work Equals Torque? Horsepower, Pulleys

Question

While reading one definition of torque, I saw its units are Newton-meter, which is the same as work. But sources usually make it a point to emphasize "even though both work and torque units are the same, they should not be confused, they are very different". One is like an object being pushed with force certain distance, the other force applied to a wrench etc. at certain length, applied around an axis of rotation.

But if we think of the pulley seen below,

Isn't radius and distance related here? If I rotate a wrench of length $r$ with force $F$ from top position to 90 degree position, isn't the same thing as pushing an object with force $F$ at a distance of $r$?

Answer 1

You can still do work by applying a torque, but that is not the same thing as torque being equivalent to work. An easy way to see this is that you can apply a constant torque, but the work done by the torque depends on the angle through which the torque rotates the object.

In your wrench example, assuming maximum torque for the given $F$ and $r$, your torque is then $\tau = rF$. If you rotate the wrench through some angle $\theta$, then the work you have done is $W=rF\theta$. This is equivalent to pushing an object a distance $r\theta$ with a constant force $F$, i.e., the "rotational work" is just the work over the arc length distance of the rotation.

Answer 2

One of the the big differences between work and torque:

Work - work is involved when a force is exerted through a distance and some component of that force is parallel to the displacement of the object that the force is acting on. The SI units of work are Newton-meters.

Torque - torque is a force whereby a component of that force is exerted at right angles to a lever arm, and that lever arm is "attempting" to rotate the object it is acting on. The SI units of torque are also Newton-meters. HOWEVER, and this is one good example of how torque differs from work - if I am changing a tire and I am putting torque on one of the lug nuts with my lug wrench in order to remove that tire, I may find that the lug nut is on so tightly that despite putting a LOT of torque on it, it refuses to move. Since there is no motion through a displacement, no work is involved, but torque is STILL being applied to that lug nut.

Answer 3

"If I rotate a wrench of length r with force F from top position to 90 degree position, isn't the same thing as pushing an object with force F at a distance of r?"

No; the distance you would push is not r, but the distance traveled by the point where you apply the force as it rotates around the center, which would be 1/42pi*r, which is greater than r.

I think we should take a step back and forget about the equations, and think intuitively about what torque and work mean. They are very different things. Force is how hard you are pushing on something, measured in Newtons. If you push a box with F force for a distance d (meters), you applied work to the box equivalent to Fd, which would have units Nm. Work is a measure of how much energy you transferred to the box.

Now, you can think of the same thing in terms of rotation, and it is completely analogous. When thinking about rotations, the analog to force is a torque; it also measures how hard you are 'pushing' or in this case how hard you are 'twisting' the wrench. It just so happens that the units of torque are also Nm, but don't let this confuse you, because here the length represents another length, not the distance traveled but the radius of rotation, so it has a different physical meaning although it shares the same units. If you multiply the torque times the angle traveled (in radians) then you get a measure of how much energy you transferred to the wrench, which is work, which will also have units of Nm, since radians is unitless.

In terms of translation:

work = how hard you push * distance traveled

work = force applied * distance traveled

N * m = N * m

In terms of rotation:

work = how hard you twist * angle traveled

work = (force applied * radius of rotation) * angle traveled

N * m = (N * m) * (unitless)

Answer 4

This is a little bit of a pet peeve of mine, so I’ll chime in with my two cents / rant.

Torque and work don’t actually share the same units, torque actually has units of $\frac{\mathrm{N\,m}}{\mathrm{rad}}$. Dimensionally speaking, the distinction is irrelevant because radians are dimensionless, however most formulas for rotational dynamics fail if you are measuring angles in degrees, rather than radians. Basically measuring torque in $\mathrm{N\,m}$ is akin to measuring angular velocity in $\mathrm{s}^{-1}$ rather than $\frac{\mathrm{rad}}{\mathrm{s}}$.

You already have an example in the answer above: work being $W = \tau\theta$ only works if $\theta$ is in radians, meaning $\tau$ itself is “intrinsically” linked to radians. Another, more long-winded, example: $U = \frac{1}{2} I \omega^2$, in order to get the correct units for energy without “dropping” radians, you would need $I$ to be measured in $\frac{\mathrm{kg\,m^2}}{\mathrm{rad}^2}$. Then you have $\tau = I \alpha$, where you can see only one of the radians cancels out, giving the aforementioned $\frac{\mathrm{N\,m}}{\mathrm{rad}}$.

How do we get this to “match” with the way torque and moment of inertia are defined in terms of force, mass and position vectors? It can be taken as a result of the presence of cross products in their definition. The usual definition is $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$ which can be re-written as $\boldsymbol{\tau} = A_\mathbf{r} \mathbf{F}$, using the cross product matrix of $\mathbf{r}$. $$A_\mathbf{r} = \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix}$$ But, you might say, there is no cross products in the definition of $I$. Well, the moment of inertia is actually a tensor quantity and, in tensor terms, is given by $I = m A_\mathbb{r} {A_\mathbb{r}}^\mathsf{T}$, two occurrences of our cross product matrix, meaning two occurrences of radians.

The same is true for angular momentum, which should have units of $\frac{\mathrm{J\,s}}{\mathrm{rad}}$.