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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The total length of a bar is 0.8m, and the pivot is exactly in
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A uniform bar of mass m 5.1kg is supported by a pivot at the bottom and...
A uniform bar of mass m 5.1kg is supported by a pivot at the bottom and a horizontal rope. The bar is length L = 2.5m and the horizontal rope is attached to the bar at a distance d = 1.975m from the bottom end. The bar makes an T angle of 0 with the vertical as shown. If the tension in the rope T = 17.8 N what angle is consistent with rotational equilibrium? Answer in degrees.
please answer all parts to each question P1300 35° Length 797 d1 pivot d2 Length Art...
please answer all parts to each
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wall length of Ladder=7.om man on middle of ladder mass of man man Ladder's center of...
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A uniform bar of mass m 5.1kg is supported by a pivot at the bottom and a horizontal rope. The bar is length L = 2.5m and the horizontal rope is attached to the bar at a distance d = 1.975m from the bottom end. The bar makes an T angle of 0 with the vertical as shown. If the tension in the rope T = 17.8 N what angle is consistent with rotational equilibrium? Answer in degrees.
please answer all parts to each
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