III. Torque for non-perpendicular forces: 1' = Fvemendimrr When the force is not perpendicular to the object it's trying to rotate, the torque it creates is equal to 1' = F 3\" sin 9. F I = Fr sin 3 where F = force r = distance from point of rotation to where F is applied r 9 = angle between the force and the object it's trying to rotate Notice what this means though: F sine is the component of the force that is perpendicular to the object. This is the component of the force that is responsible for trying to get the object to rotation, Fmrpmmmlar. """" F 1' = F I r where F J_ = component of the force perpendicular to object it's trying to rotate 1' r = distance from point of rotation to where F is applied Fi=Fsin6 Immrtant: Don't forget about the sign. Counterclockwise torque is msitive Clockwise torque is negative A. The rod in the gure can rotate about its left end. The force shown is equal to 12 N and the length of the rod is 1.5 m. a. The torque caused by this force around the left end of the rod is (circle one): negative zero positive h. Calculate the torque caused by the force F. Use Fir and don't forget to include the sign! Do NOT use Fr sin 9, you will get the question wrong! c. A second force F; = 12 N is added at the mid-point of the rod acting as shown. The torque caused by this force around the left end of the rod is (circle one) negative zero positive d. Calculate the net torque around the left end of the rod. Don't forget to include the sign! e. A third force is added as shown (perpendicular, midway between the left end and F2) to keep the rod from rotating. What must the magnitude of this force be? Tutorial: Torque and statics IV. Torque for non-perpendicular forces: 1' = Frpemendimm F 3 'l.' = Fr where F = force dJ_ = perpendicular distance from the point of rotation to the line along which the force is applied These two approaches are complementary. Depending on what's given, sometimes using F J_r is more convenient, other times using Fri. F2 A. In the situation shown, the rod (length 2.0 m) can rotate about its 300 midpoint. F1 = 10 N, F2 = 20 N. Which approach is more convenient to use to calculate the net torque in this situation? a. 1: = F J_r b. 1: = Fri F1 c. There's no advantage to picking one or the other. The torque caused by this force around the middle of the rod is (circle one): negative zero positive B. Use the approach you identied to calculate the net torque acting on the rod. C. For the situation shown, which equation can be used to calculate the torque caused by the gravitational force about the pivot indicated (bottom of the rod) on the rod of mass 8 kg? 3.. 1: = F J_r b. 1: = Fri c. There's no advantage to picking one or the other. The torque caused by the weight around the bottom end of the rod is (circle one): negative zero positive D. Use the approach you identied to calculate the torque caused by the weight. Tutorial: Torque and statics E. A second force F is added to the rod as shown. What must be the magnitude of this force in order for the rod to be in equilibrium? F. The pulley shown is horizontal and has a radius of 0.5 m. It is xed in place and can rotate about its midpoint. Two people are pulling on the rope that goes around the pulley with forces F1 = 15 N and F2 = 20 N. F Which equation can be used to calculate the net torque acting on the pulley around its center? a. 1: = F J_r b. 1: = F11 c. There's no advantage to picking one or the other. The torque caused by F1 is (circle one): F2 negative zero positive The torque caused by F2 is (circle one): negative zero positive G. Use your answers to F above to calculate the net torque acting on the pulley around its center. Don't forget about the sign