Torque and Rotational Equilibrium Name Section Torque is the rotational analog of force. If you want something to move (translation), you apply a force; if you want something to rotate, you apply a torque. Theory Torque is defined as τ = Fd () where F is a force and d a distance. This distance is known as the moment (or lever) arm of the torque and is defined as the perpendicular distance from the line of action of the force to the axis of rotation. The SI unit of torque is a Nm. By convention, torques which would produce a clockwise rotation are considered negative; torques which would produce a counterclockwise rotation are considered positive. If a system is in equilibrium, then two conditions must be met:. The sum of the forces on the system is zero.. The sum of the torques about any point must be zero. You ve already done a lot with the first of these; now let s look at the second in more detail. Notice that it states that the sum of the torques (the net torque) about any point must be zero. This means that you can calculate and sum torques about any point and you will get zero if the system is in equilibrium. However, there are a few things to keep in mind. The moment arm for each torque changes when you change the point about which you are calculating and summing torques (the axis of rotation). Also, it is possible that you will have a different number of torques about different points. Finally, a torque that would produce a clockwise rotation about one point may produce a counterclockwise rotation about another point. Look at the following example. Two people are standing on a seesaw and are in equilibrium. From condition above, F UP m g m g = Here, we are assuming that the mass of the seesaw plank is negligible compared to the weights of the people so that the upward force at the support is just the sum of their individual weights. Condition dictates that (summing torques about the support point) F d F d m gd m gd = = The upward force at the support contributes no torque about this point because d =! But condition states that the sum of the torques about any point is. What if we were to instead sum torques about the left end of the seesaw? 5 Page of 6
Here we have m gd + FUP dup m gd = and the upward force on the seesaw plank will contribute a torque about the left end of the seesaw. Look at it this way if F UP were the only force acting on the seesaw, it would rotate the plank counterclockwise about the left end. In summary, there are torques about the support point, but 3 about the left end of the seesaw. Additionally, the moment arms for the torques due to the people change since they are measured relative to different points. The one thing that both situations do have in common is that the sum of the torques (net torque) is zero. Apparatus Meter stick, Knife-edge clamp, Support, Hooked Masses, String, Scissors, Triple-beam balance. Procedure The system you assemble in each of the following procedures starts with the meter stick balanced on the support by itself. This way, the meter stick itself will contribute no torque. The hooked masses can be suspended from the meter stick with loops of string. You are asked to draw a diagram of each system you set up. I want to see a meter stick with the support and all the masses. Label the magnitude of each mass (in grams) as well as the locations of the masses and the support (in cm). These locations should be where they actually were on the meter stick; i.e., from cm cm. This will make it easier to check your calculations. I know that the SI unit of torque is a Nm, but to avoid lengthy conversions, I want torques in gcm. In other words, your forces will be in grams, and your distances in centimeters. Two Masses in Equilibrium Suspend a g mass from the meter stick on one side of the support and a g mass on the other side. Adjust the locations of these masses until equilibrium is achieved (the meter stick is balanced again). Draw a diagram of this system below, then calculate and sum the torques about the support point. Show all work. Diagram :. Net Torque about the Support Point: 5 Page of 6
Three Masses in Equilibrium Suspend a g mass and a 5g mass on one side of the support (different locations), and a g mass on the other side. Adjust the locations until the system is again balanced. Draw a diagram of this system below. Calculate and sum the torques about the support point; show all work in the space provided. Now, calculate and sum the torques about the cm point on the meter stick. Show all work in the space provided. Finally, calculate and sum the torques about a third point in the system of your choosing (other than the already tested). Make sure to specify which point you are using, and show all work in the space provided. Diagram :. Net Torque about the Support Point:. Net Torque about cm on the Meterstick: 5 Page 3 of 6
3. Net Torque about the Third Point: Point used (cm): Determining an Unknown Mass Suspend a g mass on one side of the support, and the unknown mass on the other side. Adjust locations until the system is again balanced. Draw a diagram of the system below, then use that fact that the system is in equilibrium to determine the mass of the unknown. Show all work in the space provided. Determine the mass of the unknown with the beam balance for comparative purposes. Diagram 3:. Determining the Mass of the Unknown: 5 Page 4 of 6
Analysis. For the masses in equilibrium procedure, what was the net torque about the support point? What should be the net torque about this point?. For the 3 masses in equilibrium, what was the net torque about the support point? What should the net torque be about this point? 3. How did the net torque about the other points you determined compare with the net torque about the support point in the 3 mass system? Is this surprising? Explain. 4. What was the mass of the unknown as determined with the beam balance? What is the percent difference between this and your experimentally determined mass? 5 Page 5 of 6
Pre-Lab: Torque and Rotational Equilibrium Name Section. What are the two conditions for equilibrium? Consider the system shown here:. What is the clockwise torque about the support point (gcm)? 3. What is the counterclockwise torque about the support point (gcm)? 4. Is the system in equilibrium? How do you know? 5. How many torques are there about the left end of the meter stick (cm)? 5 Page 6 of 6