# Machines pulling an object upwards, find forces

1. Mar 29, 2004

### Format

Hey guys, im workin on some stuff dealin with machines at the moment but cant really figure it out. Seems simple but the book just doesnt help.
If you have an object of 1000 N being pulled 5m upwards by a pully, and 20.0m of rope is pulled, then how do you figure out the Effort Force? I thought it was the Resistance force (1000) divided by the distance (5)? Anyone mind helpin me out? Thx

2. Mar 29, 2004

### Format

Nevermind, i seemed to of figured it out while i was typing lol. Guess i just needed to too see the question better. Thx anywayz!

3. Mar 29, 2004

### Format

k sorry for all the posts but In the same question as above, how do you figure out Work Input and Work Output?

4. Mar 29, 2004

### NateTG

They should be equal unless you want to start accounting for friction.

5. Mar 29, 2004

### Format

Well part of the question is what force is used to overcome friction if the actual effort is 300 N? But i dont know wut the friction is .

6. Mar 29, 2004

### Staff: Mentor

I presume you know the resistance force? And you can figure out the ideal effort force if the machine were frictionless? Actual effort force minus ideal effort force = force needed to overcome friction.

7. Mar 29, 2004

### Format

k well this is all i have:

Stan raises a 1000-N piano a distance of 5.00 m using a set of pulleys. Stan pulls in 20.0 m if rope.
a) How much effort does stan apply if this was an adeal machine? - Done
b) What force is used to overcome friction if the actual effort is 300 N ?
c) What is the work output?
d) What is the work input?

Ive tried but cant really get it. Book doesnt help, and i dont have anyone to ask. So any help would be appriciated! Thx

8. Mar 29, 2004

### Janitor

Clue on b

He's lifting an object with the help of a 4:1 mechanical advantage. So the force he himself applies would ideally be 1000 divided by 4, right? So when you've done that division, compare the number you get to 300, and that will tell you how much force is going into overcoming real-world friction.

9. Mar 29, 2004

### Format

k figured that out, thx! Still confused with The work input/output however.

10. Mar 29, 2004

### Janitor

Work output = force applied directly to the object (no friction!) times the height it is lifted

Work input = force the person applies to the rope (friction included!) times the length of rope he pulls

11. Apr 7, 2004

### PhysHey

Regarding the work input, is the force that Stan applies 300 N? And the length of rope he pulls 20 m?
That works out to be the answer when multiplied. (Win= E x De - displacement). That's what I have from my notes. I get the correct answer but I'm not sure on the concept. How do you know it's the 20 m of rope and not the "distance of 5.00 m?.....hmm

12. Apr 8, 2004

### Staff: Mentor

Yes.
The work input is the applied effort force times the displacement over which that force acts, which is 20 m. When the object is displaced by a distance of 5.0 m, that displacement is due to a different force--the output force.