At one point during its swing, a wrecking ball exerts a tension force of F_T = (

Question

At one point during its swing, a wrecking ball exerts a tension force of F_T = (800 i - 600 j) N on its cable, which makes an angle alpha = 36.9 degree with the horizontal. The crane's 9-m-long boom is at an angle of theta = 53.1 degree with the horizontal. With point P as the origin of the coordinate system and the horizontal is the x-axis, the tension force F_T is applied to a point with position vector r = (5.4 i + 7.2j) m as shown in the figure below. What is the magnitude of the torque exerted by the wrecking ball on the crane about the axis passing through point P (the origin)? Use sin(36.9) = 0.6, cos(36.9) = 0.8, sin(53.1) = 0.8, and cos(53.1) = 0.6. 9000 N m 17,640 N m 2520 N m 5760 N m 3240 N m

Explanation / Answer

a) 9000 N.m

from the given data, |FT| = sqrt(800^2 + 600^2)

= 1000 N

torque, T = r cross FT

= r*FT*sin(theta)

= 9*1000*sin(180 - 36.9 - 53.1)

= 9000 N.m

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