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Unformatted text preview: Here’s the tube (2.4 meters long) with its load: 40 kg 60 kg __________X__________M 0 m 1.2 m 2.4 m (40 kg) (1.2 m) + (60 kg) (2.4 m) Dist of CG from 0 meter end is = ______________________________ = 1.92 m (40 kg + 60 kg) Again, I wouldn’t ask you to calculate this, but I’d like you to have a general feel for where the CG of such a contraption lies. 16. The torque about some axis due to an applied force exerted at a distance from the axis is equal to the component of that force perpendicular to the lever arm connecting the force’s point of contact to the axis times the length of the lever arm. (continued….) For two people balanced on a seesaw, one exerts a clockwise torque about the fulcrum which is balanced by the other person’s equal and opposite (counterclockwise) torque, producing a net torque of zero about the fulcrum....
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This note was uploaded on 10/19/2011 for the course PHY 2020 taught by Professor Staff during the Spring '08 term at University of Florida.
- Spring '08