Posted: 1/22/2015 8:29:07 PM EDT
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Question: If you can only exert 70 pounds of force and you need to life a 160-pound steel drum up 2 feet, what is the minimum length of ramp you should set up?
foot-pounds / length of ramp = force 160 pounds up 2 feet is 320 foot-pounds. 320 / x = 70 answer = just under 4' 7", therefore 4' 7" is the minimum length of the ramp. Did I do it right? |
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Quoted:
Question: If you can only exert 70 pounds of force and you need to life a 160-pound steel drum up 2 feet, what is the minimum length of ramp you should set up? foot-pounds / length of ramp = force 160 pounds up 2 feet is 320 foot-pounds. 320 / x = 70 answer = just under 4' 7", therefore 4' 7" is the minimum length of the ramp. Did I do it right? Yes. The work is force X distance, so you need to move the drum far enough on a ramp to add 160(2) ft-lbf of energy (work). (The weight of the drum is a force.) |
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Quoted:
Using trig, find the angle that produces a 70 lbs force down the ramp. Then with that angle and 2 ft height find the length of the ramp using trig. The downward acting force on the ramp is 160 pounds. It could be resolved into two forces, one parallel to a ramp and the other perpendicular to the ramp, but it won't help solve this problem. This is a work - energy problem, not trig. The drum needs be moved 2 feet vertically. So its energy has to be increased by 160(2) foot pounds. The amount of work to raise the drum over a longer distance on a ramp is Work = Force(Distance) = Weight(Vertical Distance). You're right, both methods yield the same angle or ramp length, but the energy method is simple. |
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Quoted:
Question: If you can only exert 70 pounds of force and you need to life a 160-pound steel drum up 2 feet, what is the minimum length of ramp you should set up? foot-pounds / length of ramp = force 160 pounds up 2 feet is 320 foot-pounds. 320 / x = 70 answer = just under 4' 7", therefore 4' 7" is the minimum length of the ramp. Did I do it right? Yer math is all kinds of fucked up! 
Here: 14,459' to the local Home Depot $20 Two messicans 14,459' back to shop 14,459' to Home Depot Total, $23 The answer is 23 
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Quoted:
The downward acting force on the ramp is 160 pounds. It could be resolved into two forces, one parallel to a ramp and the other perpendicular to the ramp, but it won't help solve this problem. This is a work - energy problem, not trig. The drum needs be moved 2 feet vertically. So its energy has to be increased by 160(2) foot pounds. The amount of work to raise the drum over a longer distance on a ramp is Work = Force(Distance) = Weight(Vertical Distance). Quoted:
Quoted:
Using trig, find the angle that produces a 70 lbs force down the ramp. Then with that angle and 2 ft height find the length of the ramp using trig. The downward acting force on the ramp is 160 pounds. It could be resolved into two forces, one parallel to a ramp and the other perpendicular to the ramp, but it won't help solve this problem. This is a work - energy problem, not trig. The drum needs be moved 2 feet vertically. So its energy has to be increased by 160(2) foot pounds. The amount of work to raise the drum over a longer distance on a ramp is Work = Force(Distance) = Weight(Vertical Distance). By "down the ramp" I meant the force vector parallel to the ramp, not the downward force. By all means you can solve it with trig. Assuming this is a frictionless problem. The approach you guys agree on is making me scratch my head but I haven't put much thought into it. |
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Quoted:
By "down the ramp" I meant the force vector parallel to the ramp, not the downward force. By all means you can solve it with trig. Assuming this is a frictionless problem. The approach you guys agree on is making me scratch my head but I haven't put much thought into it. Quoted:
Quoted:
Quoted:
Using trig, find the angle that produces a 70 lbs force down the ramp. Then with that angle and 2 ft height find the length of the ramp using trig. The downward acting force on the ramp is 160 pounds. It could be resolved into two forces, one parallel to a ramp and the other perpendicular to the ramp, but it won't help solve this problem. This is a work - energy problem, not trig. The drum needs be moved 2 feet vertically. So its energy has to be increased by 160(2) foot pounds. The amount of work to raise the drum over a longer distance on a ramp is Work = Force(Distance) = Weight(Vertical Distance). By "down the ramp" I meant the force vector parallel to the ramp, not the downward force. By all means you can solve it with trig. Assuming this is a frictionless problem. The approach you guys agree on is making me scratch my head but I haven't put much thought into it. I agree with your solution, just not the method. I added a correction above. 160sin(angle) = 70 => angle = 25.844 deg (Ramp length)sin(angle) = 2, ramp length = 4.57 feet. |