Solve Pulleys & Mass Homework: Lift 60N Box w/ 3 Pulleys

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A man is attempting to lift a 60N box using a system of three pulleys and two ropes. The discussion revolves around calculating the force required for the man to pull on the first rope, with considerations for static equilibrium and the tensions in the ropes. After analyzing the forces and drawing free body diagrams (FBDs), it is determined that the tension in the green rope is 24N, while the force the man must exert is 12N. The participants clarify that the tensions are not the same across the ropes and emphasize the importance of correctly labeling and solving the equations for each pulley. Ultimately, the man needs to pull with a force greater than 12N to successfully lift the box.
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Homework Statement


A man is trying to lift a box using three pulleys and two peices of rope. Each pulley can turn without friction and has negligible mass.The first rope, attached to the man, paases down under pulley A, up over pulley B and down to the frame of pulley C. The second rope is tied to the frame of pulley A and passes down under pulley C and up to the ceiling. If the box weighs 60N with what force must the man pull on the first rope to lift the box?


Homework Equations





The Attempt at a Solution


To start off i know i have to draw an FBD. but would there be 3 FBDs for the pulleys or just one for the mass?
 

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The ropes are wrapped around in such a way that just one FBD won't help you.
 
If you assume that the setup is in static equilibrium, then given that the pulleys and ropes are massless and frictionless you should be able to label the tensions in all the rope segments with a bit of careful thought. Start by assigning the unknown tension T to the rope held by the person (the one you've made red in your diagram). Can you determine the tension in the other rope?
 
so would the tension be the same for both ropes and also when i am finding the tension I am getting a value of zero?
 
Loading said:
so would the tension be the same for both ropes and also when i am finding the tension I am getting a value of zero?

No it wouldn't be the same.
 
ok so i would have an equation for each the ropes then...but when i am calculating tension i am getting zero...i thinmk my FBD equation is wrong. For the green rope would it be Fnet=mg + 3T
 
Loading said:
ok so i would have an equation for each the ropes then...but when i am calculating tension i am getting zero...i thinmk my FBD equation is wrong. For the green rope would it be Fnet=mg + 3T

Where did you get 3T from? And what exactly is Fnet?
 
the net force or mass x acceleration and i think that's where i am wrong ..would it be 2T instead?
 
What if you considered each pulley and found a relation between the tensions?
 
  • #10
so draw an FBD for each pulley and then try to find tension?
im assuming the tension in pulley A and B will be equal?
 
  • #11
Yeah do that! And yes, the tensions in the ropes of both pulleys are same, since they are friction less.
 
  • #12
ok so here's what i got
for pulley A i have 2 tensions pulling up and one pulling down
for B i got on tension pulling up and two pulling down
i don't know if that's right and also how would i do C
 
  • #13
I suggest you only focus on pulleys A and C. The FBD of pulley B is not necessary.

EDIT: If you can talk about the tensions for pulley A, why can't you do the same for C? Post some equations that you can write.
 
  • #14
ok so pulley A will have 2 tensions up and one tension down
how would i do pulley C?
would it also have 2 tensions pulling up and the mass pulling down?
 
  • #15
Loading said:
would it also have 2 tensions pulling up and the mass pulling down?

Yep.
 
  • #16
ok so i got 2 equations:
1. ma= T
2. ma = 2T - mg

would i now add botht these equations together?
 
  • #17
There are 2 different tensions in each of the ropes.

There is no acceleration because of what gneill said.

Before you said that for pulley A, there were 2 tensions up and 1 down, and same for C. Label the tensions for each color and put them in equations and solve for both.

Also, use 60 N instead of mg.
 
  • #18
yeah but that's the thing for A i am getting the tension to be zero.
The equation i have for A is ma=t is this right .
Also for pulley C i got 30N
 
  • #19
Loading said:
Also for pulley C i got 30N
edit: this is not correct, because there are 3 tensions, 2 of which are the same.
Loading said:
The equation i have for A is ma=t is this right .
No it's not. a=0, and there are 3 tensions, 2 of which are the same.
 
  • #20
ok i got 15N...would this be the force the man would use to pull the box ?
Also why did we not use pulley B?
 
  • #21
Sorry I edited my post above, so 15 N is wrong. The diagram in my head was different.

Pulley B is just stationary and is acted upon by another force apart from the 2 different tensions, so it just gives us another unneeded unknown.
 
  • #22
ahh ok so with the new ones i got the tension for pulley C 24N and Pulley A to be 15N.
now would 15N be the force that the man must pull with?
 
  • #23
Loading said:
ahh ok so with the new ones i got the tension for pulley C 24N and Pulley A to be 15N.
now would 15N be the force that the man must pull with?

24 N is correct and this is the tension in the green rope. 15 N is not correct. Check your equations..

The tension in the red rope is the force that the man must pull it with.
 
  • #24
sorry i meant to write 12N..my bad
and this would be the force the man would have to use right
 
  • #25
Loading said:
sorry i meant to write 12N..my bad
and this would be the force the man would have to use right

Yes. :smile: The force must be greater than 12 N to lift it up.
 
  • #26
Yes i understand..ok Thanks a lot u really walked me through this question.
Thank u sir Have a good day :)
 
  • #27
You're welcome! :smile:
 

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