The Physics Philes, lesson 51: Pipe Dreams
Last week we started discussing a brand-spanking-new topic: torque! At the end of last week’s saga, I left you with several different methods of calculating torque. This week I’ll do my best to show you how those equations work.
Are you ready to go? LET’S DO IT.
We have a plumber uses a “cheater” to help loosen a pipe fitting. (A cheater is basically a piece of scape pipe over the handle of a wrench.) The plumber puts his full weight into it – all 900 N. The distance from the center of the pipe to the point where the weight acts is 0.80 m. The handle and cheater make a 19 degree angle with the horizontal. Find the magnitude of the torque.
First, let’s draw a diagram.
Whoa! Look at all that information/needed information! I color-coded it for easy reference. (Also, it’s pretty.)
- The red line is the distance from center of the pipe fitting to where the force is applied; 0.80 m.
- The solid blue line is the force applied; 900 N.
- The blue dotted lines are the tangential and radial force components.
- The green line is lever arm.
- The black line is the horizontal, from which the angle to the wrench handle is measured.
Hopefully now the picture makes a little more sense. Now, let’s use the three equations introduced last week to figure out the magnitude of the torque.
To use the first equation for torque (τ = Fl), we need to find for l, the green lever arm. To do this, we use sine, i.e. the opposite side of the triangle divided by the hypotenuse. In this case, the opposite side is l and the hypotenuse is r = 0.80 m. So the equation we get is:
(By the way, sin 109 and sin 71 should be in degrees, but I’m using LaTeX for the first time and I’m still feeling my way around it. Also, since I don’t really know what it’s supposed to look like, let me know if it looks all wonky!)
Now that we know the value of the the lever arm we can use τ = Fl to find the torque!
All right! 680 N·m worth of torque! Now let’s use the second of our torque equations to see if we can get the same answer. That equation, if you will remember, is τ = rFsinφ. (This should make sense, as we’ve already established that l equals rsinφ.) The angle φ is measured between the vectors r and F. You can see from our diagram that the angle φ is 109 degrees. So all we have to do is plug in the numbers.
Interesting. Once again, we came up with 680 N·m. Now, we have one more equation to try. And it will take a little explanation.
Look back at our diagram for a second. The solid blue line represents not only the 900 N of force applied to the wrench, but also the force vector. As we learned ever so long ago, that force vector has two components. In torque problems we call those components the tangential and the radial components. The radial component is parallel to the vector r and the tangential component is perpendicular to vector r. As we learned last week, the radial component produces no torque with respect to point O because its lever arm with respect to that point is zero. The tangential component, on the other hand, is equal to Fsinφ. If we just do a little substitution we find that τ = rF_tan (or F_tan·r, but I thought the former was easier to read in this format.)
Now let’s apply this to the problem. The tangential component of F acts perpendicular to r. Since vector r is oriented 19 degrees from the horizontal, that means the perpendicular to r – in this case F_tan – is oriented 19 degrees from the vertical, which in this case is F. This all means that F_tan = F(cos 19). So, solving for F_tan:
Now we can just slip 851 N into τ = rF_tan:
680 N·m again! All the equations work! Woo hoo! This was fun, right? Well I had fun, at least.
Let me know if anything is out of place, or weird, or wrong. Next week we’ll start a discussion on the relationship between the angular acceleration of a rigid body and torque. Squee! I can hardly wait!
Featured image credit: Flickr