Answer in units of N · m

Question

The arm of a crane at a construction site is 17.0 m long, and it makes an angle of 14.2 ◦ with the horizontal. Assume that the maximum load the crane can handle is limited by the amount of torque the load produces around the base of the arm. What maximum torque can the crane withstand if the maximum load the crane can handle is 450 N? Answer in units of N · m

Answer 1

1) Start with drawing a picture. You have a fixed base where the arm resists all moment. You have a free end that the load rests at the extreme edge on from the base. The arm is elevated 14.2 degrees above the horizontal, and the load due to gravity will point exactly downward. This load, Fmax = 450 N. The arm length is 17 m at 14.2 degrees.

2) Set up a sum of all moments equation. E M = M max (N*M) =< Fmax * Arm length * cos(angle of elevation). As the arm is elevated above the horizon the effective length of the "lever" generating moment at the crane base is actually shorter than the length of the arm. You can reduce the length of arm using trigonometry. On a right triangle like the one formed here you can use the cosine function to reduce the length and find the horizontal distance. Take care not to use the sine function, as you will get the vertical distance and an incorrect answer. Also take care to use your calculator in degree and not radian mode.

3) Enter real values into your equation, M max =< 450 N * 17 M * COS(14.7 degrees). When simplified, you get 7416.26 N*M or 7.42 kN*M as the torque capacity. You can check your work using the units test, where you verify that you multiplied a length (M) by a force (N) and received moment (N*M).

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