Example of Torque

1. Example of Torque Consider two people carrying a board with a weight on it – see the diagram below. The weight is 100 lb, and is located 7 feet away from the left end. The board is 8 feet long and we neglect the weight of the board. How much force does each person have to exert? FL = ? FR = ? W = 100 lb 7 feet 8 feet

2. Example of Torque We recognize this as a statics problem, and a one dimensional problem (in y). This gives us one relation: SF = 0. We identify the force the left person exerts, FL, the force the right person exerts, FR, and the Weight. SF = +FL + FR – 100 lb = 0 . FL = ? FR = ? W = 100 lb 7 feet 8 feet

3. Example of Torque SF = +FL + FR – 100 lb = 0 . This is one equation in two unknowns. There are lots of ways of dividing up the weight among the left and right persons. What does determine how much each has to exert? FL = ? FR = ? W = 100 lb 7 feet 8 feet

4. Example of Torque SF = +FL + FR – 100 lb = 0 . What is important is where the weight is placed! To get our second equation, we recognize this as a torque situation: St = 0 where each of the three forces has a distance (moment arm, r). But where do we measure each moment arm from – since there is no actual rotation about any point? FL = ? FR = ? W = 100 lb 7 feet 8 feet

5. Example of Torque SF = +FL + FR – 100 lb = 0 . Since there is no obvious point about which the board rotates, we are free to choose any point. Since the diagram indicates the distances measured from the left end, we can initially choose that point. Also for definiteness sake, let’s choose a counter-clockwise (CCW) rotation as being positive. FL = ? FR = ? W = 100 lb 7 feet 8 feet

6. Example of Torque SF = +FL + FR – 100 lb = 0 . From the choices of the previous slide, we see that rL = 0 ft, rR = 8 ft, and rW = 7 ft; and the torque due to FL is zero, the torque due to FR is CCW and so is positive, and the torque due to W is CW and so is negative: St = {FL * 0 ft} + {FR * 8 ft} - (100 lb * 7 ft) = 0. FL = ? FR = ? W = 100 lb 7 feet 8 feet

7. Example of Torque SF = +FL + FR – 100 lb = 0 . St = {FL * 0 ft} + {FR * 8 ft} - (100 lb * 7 ft) = 0. We now have two equations for two unknowns, and we can solve the problem. From the torque equation we can solve for FR: FR = (100 lb * 7 ft) / 8 ft = 87.5 lb. And from the force equation: FL= 100 lb – 87.5 lb = 12.5 lb. FL = ? FR = ? W = 100 lb 7 feet 8 feet

8. Example of Torque FR = (100 lb * 7 ft) / 8 ft = 87.5 lb, andFL= 100 lb – 87.5 lb = 12.5 lb. Note that the further the weight is from the person, the lighter is the necessary force. Also note that by choosing the measuring point at the left force location, we effectively eliminated the FL unknown from the torque equation making it easier to solve. FL = ? FR = ? W = 100 lb 7 feet 8 feet

9. Example: YoYo Consider a yoyo resting on a horizontal surface. The string is wound so that it comes out horizontally on the lower side and is pulled with a tension, T. Will the yoyo come toward you when you pull the string, or will it go away? T

10. Example: YoYo How do we figure this out? We use Newton’s 2nd Law for the forces, and then use the extension of that law for torques: ΣFx = m*ax; ΣFy = 0; Σt = I*a . We need to identify the forces, and we need to determine where those forces act. T

11. Example: YoYo ΣFx = m*ax; ΣFy = 0; Σt = I*a . Gravity (W=mg) acts down and acts at the center of mass. The Contact Force (Fc) acts up and acts at the surface. The Tension (T) acts to the right and acts at the base of the inner radius, r. Friction (Ff) acts to oppose the yoyo sliding towards the right, so it is directed to the left and acts at the surface at a distance of the outer radius, R. T

12. Example: YoYo ΣFx = T – Ff = m*ax ΣFy = Fc – W = 0 Σt = - T*r + Ff*R = I*a . If ax and a are positive (yoyo going to the right), then T > Ff and Ff*R > T*r - this is possible since R > r. If ax and a are negative (yoyo going to the left), then Ff > T and Ff*R < T*r - this is not possible since R > r. T