The body in the figure is pivoted at O. Three forces act on it in the directions
ID: 1910084 • Letter: T
Question
The body in the figure is pivoted at O. Three forces act on it in the directions shown: FA = 14 N at point A, 6.7 m from O; FB = 15 N at point B, 4.1 m from O; and FC = 19 N at point C, 3.1 m from O. Taking the clockwise direction to be negative, what is the net torque about O?
The body in the figure is pivoted at O. Three forces act on it in the directions shown: FA = 14 N at point A, 6.7 m from O; FB = 15 N at point B, 4.1 m from O; and FC = 19 N at point C, 3.1 m from O. Taking the clockwise direction to be negative, what is the net torque about O?Explanation / Answer
perpendicalur distance of FA from O is 6.7 cos 45 = 4.737 m & torque is anticlockwise(+ve)
perpendicalur distance of FB from O is 4.1 m & torque is clockwise(-ve)
perpendicalur distance of FC from O is 3.1 cos 70 = 1.06 m & torque is anticlockwise (+ve)
SO, net torque is
14(4.737) - 15(4.1) + 19(1.06) = 24.968 N-m (clockwise)
Therefore, the answer is 24.968 N-m (clockwise)
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