When tightening bolts in your car, you see that the nuts must be lightened to 12

Question

When tightening bolts in your car, you see that the nuts must be lightened to 120 foot-pounds of torque. The foot-pound is the imperial unit of torque: 1 foot - 0 3048 m; 1 pound = 4.448 N. If you are using a wrench that is 60 cm long, how much force do you need to apply to the wrench? (270 N) At the gym, you are doing leg raises as shown in the diagram below. Find the torque of the weights (assume you're lifting at constant speed) the muscle torque the net torque (and direction) (35 Nm; 48 Nm; 13 Nm) Inside the engine of a car, the combustion inside a cylinder produces 2500. N of force in the push rod. Find the torque on the crankshaft. (125 Nm)

Explanation / Answer

(5) Muscle force Fm = 2200 N        and the distance from pivot x1 = 2.2 cm       Mass of the weights m = 5.5 kg     And the tension acting at distacne from pivot x2 = 0.65 m (a) Torque of the weights                 = F x2                   = mg x2                  = (5.5 kg) (9.8 m/s2) (0.65 m)                  = 35 N.m (b) Muscle torque                 = Fm x1                   = (2200 N) (0.022 m)                  = 48 N.m (c) Therefore the net torque about pivot                = 48 Nm - 35 Nm                     = 13 Nm (5) Muscle force Fm = 2200 N        and the distance from pivot x1 = 2.2 cm       Mass of the weights m = 5.5 kg     And the tension acting at distacne from pivot x2 = 0.65 m (a) Torque of the weights                 = F x2                   = mg x2                  = (5.5 kg) (9.8 m/s2) (0.65 m)                  = 35 N.m (b) Muscle torque                 = Fm x1                   = (2200 N) (0.022 m)                  = 48 N.m (c) Therefore the net torque about pivot                = 48 Nm - 35 Nm                     = 13 Nm (5) Muscle force Fm = 2200 N        and the distance from pivot x1 = 2.2 cm       Mass of the weights m = 5.5 kg     And the tension acting at distacne from pivot x2 = 0.65 m (a) Torque of the weights                 = F x2                   = mg x2                  = (5.5 kg) (9.8 m/s2) (0.65 m)                  = 35 N.m (b) Muscle torque                 = Fm x1                   = (2200 N) (0.022 m)                  = 48 N.m (c) Therefore the net torque about pivot                = 48 Nm - 35 Nm                     = 13 Nm                   = (2200 N) (0.022 m)                  = 48 N.m (c) Therefore the net torque about pivot                = 48 Nm - 35 Nm                     = 13 Nm

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