I‘m doing homework, and my workbook only provides instructions on how to find the weight. I‘m assuming I need to reverse some operations, but I‘m clueless as to how to go about this. Can anyone explain to me how to solve this? The question is:Diameter of axle 3.5Axis of handle 21Weight lifted 180 lb.Force ?
Easy. The TORQUE is what you're going to have to equate. Torque of force on the handle Touque of whieght on the axle. Torque equals the magnitude of the force that acts perpendicular to teh axis of the axle (Lbs is a measure of force, not of mass) times the distance between the axis of the axle and the point the force is applied, which would be the radius of the axle for the weight and the lenghth of the handle for the force you're looking for. 1.75 X 180 21 X Force Force 1.75 X 180 / 21
just reverse what your workbook says. probably something along the lines of tF*r. you have to different forces: the force on the handle and the weight lifted. r is the radius or distance from the center of the axle, so the axis of the handle and half the diameter of the axle. the torque will be the same for both, so you can set up an equality with everything: F*21t180lb*3.5/2 F*21t180lb*1.75/2 then solve for F with using the known quantities. F180lb*1.75/21 F15lb Force 15 lb.
