the arm of a crane is 17.0 m long and makes an angle of18 o with the horizontal.

Question

the arm of a crane is 17.0 m long and makes an angle of18o with the horizontal. assume that the maximumload for the crane is limited by the amount of torque theload produces around the base of the arm. a) what is the maximum torque the crane can withstand if themaximum load is 454 N? (Nm) b) what is the maximum load for this crane at an angle of44o? (N) the arm of a crane is 17.0 m long and makes an angle of18o with the horizontal. assume that the maximumload for the crane is limited by the amount of torque theload produces around the base of the arm. a) what is the maximum torque the crane can withstand if themaximum load is 454 N? (Nm) b) what is the maximum load for this crane at an angle of44o? (N)

Explanation / Answer

Modified: --------- length L 17 m angle = 18 degrees (a). maximum load  F =454 N we know Torque T = FL sin For maximum torque = 90 degrees So, maximum torque T = FL                                   = 454 N * 17 m                                   = 7718 N m (b). maximm load F ' = T / ( L sin 44 )                                  = 7718 / ( 17 * 0.6946)                                  = 653.55 N

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