15.6) A uniform horizontal lever of length 2 m is pivoted at its mid-point. The

Question

15.6) A uniform horizontal lever of length 2 m is pivoted at its mid-point. The following forces are acting on the lever: a 3 N upward force (force with upward vertical component) that makes an angle of 40 deg with the horizontal-left acting at the right end of the lever, a 8 N vertically downward force acting at the left end of the lever, and a 0.73 N vertically downward force acting at a point on the lever 0.2 m away from the right end. Calculate the net torque acting on the lever.

The correct answer is 9.344 N m, but I need to know how to get there (formulas used, each step taken). Thank you!

Explanation / Answer

torque to a force F about an axis is given as :

torque = r x F = rFsin@

(where r is distaplacement vector of point of application of force from the axis and @ is the
angle between r and F )

for 3N force, r = 1m and @ = 40 deg

torque1 = 1 x 3 x sin40 = 1.928 Nm anticlockwise ( direction is found by putting you hand in the direction
of r vector and then folding fingers in the direction of force , then direction of thumb will give you
the direction of torque)

for 8N force, r = 1m and @ = 90 deg

torque2 = 1 x 8 x sin90 = 8 Nm anticlockwise

for 0.73N, r = (1-2) = 0.8 m and @ = 90 deg

torque3 = 0.8 x 0.73 x sin90 = 0.584 Nm clockwise.

Net torque = torque1 + torque2 + torque3 = 1.928 + 8 - 0.584 = 9.344 Nm anticlockwise

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