Science Olympiad Student Center
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labchick wrote:A 100 meter long first-class lever has a load of of 2 kilograms 75 meters away from the fulcrum. You are applying a force of 15 newtons and the lever is in equilibrium. What is the efficiency of the machine?
AMA=19.6/15=1.3066 IMA=25/75=0.3333 Efficiency=AMA/IMA=1000% (with significant figures) If, however, you mean 2 N force: AMA=2/15=0.1333 IMA=25/75=0.3333 Efficiency=AMA/IMA=40% (with significant figures)
Yes, sorry, I did mean 2 N force not 2 kg. Oops. Your answer is correct. Your turn to ask a question!Unome wrote:labchick wrote:A 100 meter long first-class lever has a load of of 2 kilograms 75 meters away from the fulcrum. You are applying a force of 15 newtons and the lever is in equilibrium. What is the efficiency of the machine?AMA=19.6/15=1.3066 IMA=25/75=0.3333 Efficiency=AMA/IMA=1000% (with significant figures) If, however, you mean 2 N force: AMA=2/15=0.1333 IMA=25/75=0.3333 Efficiency=AMA/IMA=40% (with significant figures)
Unome wrote:A person with a mass of 25 kg stands in the center of the beam and walks towards the right (from your point of view). The lever bar weighs 6 kg, distributed evenly. Past what distance from the edge of the bar will the lever tip over?
The lever is 6.4m in length, so each meter weighs 0.9375, or 15/16, kg. This means, on the right side of the right fulcrum, 33/32 kg, and on the left, 159/32 kg. This is distributed evenly, so it can be thought of as a single weight of that quantity in the middle of a massless lever arm, making the clockwise torque 363/640 kg•m and counterclockwise torque 8427/640 kg•m. Subtracting, that's a total of 8064/640, or (simplifying) 63/5, kg•m counterclockwise. Therefore, since the lever's in equilibrium, 25 kg * d meters=63/5 kg•m, so d=63/125 m, or 0.504 m. Therefore, the person can walk [i][b]0.504 m[/b][/i] to the right of the rightmost fulcrum before the lever tips; with significant figures, that's 0.5 meters, or 0.50 if you meant to make the lever weight 6.0 kg.
From team member (not myself as the coach):mjcox2000 wrote:In the image of the pulley system shown below, where there are 2 concentric circles over 2 intersecting ropes, those ropes are tied together at that point; triangles in the image symbolize fixed anchor points. How much downwards effort force must be applied to the arrow (on the rightmost rope) to lift the 10kg weight, as illustrated? Assume that this pulley system is an ideal machine, and g=9.81 m/s^2.
9.81N
Correct! Your turn to post a question.JonB wrote:From team member (not myself as the coach):mjcox2000 wrote:In the image of the pulley system shown below, where there are 2 concentric circles over 2 intersecting ropes, those ropes are tied together at that point; triangles in the image symbolize fixed anchor points. How much downwards effort force must be applied to the arrow (on the rightmost rope) to lift the 10kg weight, as illustrated? Assume that this pulley system is an ideal machine, and g=9.81 m/s^2.
9.81N
Would you be able to explain this solution? Really complex pulley systems are one of the things I'm currently having trouble with.JonB wrote:From team member (not myself as the coach):mjcox2000 wrote:In the image of the pulley system shown below, where there are 2 concentric circles over 2 intersecting ropes, those ropes are tied together at that point; triangles in the image symbolize fixed anchor points. How much downwards effort force must be applied to the arrow (on the rightmost rope) to lift the 10kg weight, as illustrated? Assume that this pulley system is an ideal machine, and g=9.81 m/s^2.
9.81N