Consider a force F = 80 N applied to a beam as shown in Fig. 8-37. The length of the beam is = 5.0 m,and = 37, so that x = 3.0 m and y = 4.

Consider a force F = 80 N applied to a beam as shown in Fig. 8–37. The length of the beam is  = 5.0 m,and  = 37°, so that x = 3.0 m and y = 4.0 m. Of the following expressions, which ones give the correcttorque produced by the force F around point P?(a) 80 N. (b) (80 N)(5.0 m). (c) (80 N)(5.0 m)(sin 37°).(d) (80 N)(4.0 m) (e) (80 N)(3.0 m). (f) (48 N)(5.0 m).(g) (48 N)(4.0 m)(sin 37°).

The torque produced by the applied force is, Force perpendicular distance. Fx 80 N 3.0 m Alternative method is,The torque produced by the applied force is, Force perpendicular distance. F L sin 80...