The total length of a bar is 0.8m, and the pivot is exactly in the middle....
The total length of a bar is 0.8m, and the pivot is exactly in the middle. Force F1 is applied at the bottom of the bar (10 N and 60 degrees from the horizontal). Find the torque including the direction by Force 1.
Homework Answers
assuming forse is towords downword 60degree from horizontal and is in 4 th quadrent
so torque about the pivot = Fcos(60) * L/2 = 10 * 0.5 * 0.4 = 2 Nm in anticlockwise direction
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