1 ft = 12 inch
60 inch = (60/12) = 5 ft
(1500 pounds) x (5 ft) = 7,500 foot-pounds
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To calculate the foot-pounds required to lift a weight, multiply the weight in pounds by the distance in feet. In this case, lifting 1500 pounds a distance of 60 inches (which is 5 feet) would require 1500 pounds * 5 feet = 7500 foot-pounds of work.
The question was going along so nicely there, until it got to the very end, and thenshot itself in the foot with "foot pounds"."Foot pounds" is a torque, not a force.-- If you apply 50 pounds of force at the end of the crank, that 50 pounds becomes175 pounds of force at the surface of the 8-inch cylinder, because the mechanicaladvantage of the crank/cylinder arrangement is 28/8 = 3.5 .-- If you apply 50 foot-pounds of torque at the end of the crank, then the force is50 ft-lb/28 inches = 50 lb x 12 in/28 in = 21 3/7 pounds .Now, the same mechanical advantage of 3.5 gives you 75 pounds at the hub.What a difference a unit makes !
80 lbs until past 45 degrees
1 horsepower = 550 foot-pounds per second 30 inches = 2.5 feet (2.5) x (220)/10 = 550 foot pounds per 10 seconds = 1/10th of a horsepower.
Assuming the pivot is positioned at a 90-degree angle and ignoring friction, it would take a 100-pound force applied at the end of a 1-foot-long lever arm to lift 100 pounds. This is based on the principle of torque, where the force applied multiplied by the distance from the pivot point determines the weight that can be lifted.
Macho Matt can lift 200 kilograms with ease. How much is this in pounds?