Simple Machines B/Compound Machines C

[labchick]

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?

[Unome]

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?

If it's implied that the load and effort force are at the very edges of the bar, and the force of gravity is assumed to be 9.8 m/s²... (Click to Show Hidden Text)

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)

[labchick]

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?

If it's implied that the load and effort force are at the very edges of the bar, and the force of gravity is assumed to be 9.8 m/s²... (Click to Show Hidden Text)

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]

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?

[mjcox2000]

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?

I don't think I know what you mean by "past what distance from the edge of the bar", but I'll try solving it anyways (Click to Show Hidden Text)

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.

[Unome]

Correct (I made a typo there; I was trying to say "past what distance from the rightmost fulcrum")

[mjcox2000]

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.

Complex pulley system.jpg

[JonB]

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.

Complex pulley system.jpg

From team member (not myself as the coach):

solution (Click to Show Hidden Text)

9.81N

[mjcox2000]

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.

Complex pulley system.jpg

From team member (not myself as the coach):

solution (Click to Show Hidden Text)

9.81N

Correct! Your turn to post a question.

[blindmewithscience]

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.

Complex pulley system.jpg

From team member (not myself as the coach):

solution (Click to Show Hidden Text)

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.