Solving a Tricky Pulley Problem: Input Forces for 200 N Load

Click For Summary

Homework Help Overview

The discussion revolves around determining the input forces required to lift a 200 N load using a system of pulleys. The original poster expresses confusion regarding the mechanics of two movable pulleys that move in opposite directions, which complicates their understanding of the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting up equations for equilibrium and the distribution of tension among the cables. Questions arise about whether tension is equally distributed and how to account for multiple pulleys in the analysis.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations of the pulley system. Some have begun to set up equations and label tensions, while others are clarifying their understanding of mechanical advantage and equilibrium conditions. There is a recognition of the complexity introduced by multiple pulleys and cables.

Contextual Notes

Participants note assumptions such as the pulleys being massless and frictionless, which may influence their calculations and reasoning. There is also mention of the challenge in visualizing and diagramming the forces involved in the system.

NanoChrisK
Messages
14
Reaction score
0

Homework Statement



"What are the input forces needed to lift the 200 N load for the arrangement shown?"

Notice how there are two movable pulleys that move in opposite directions as the weight is lifted. That is the confusing part to me.

Homework Equations



Ma=Fo/Fi

The Attempt at a Solution



Attempt: Ma=Fo/Fi

Fi=Fo/Ma

Fi=200 N/3

Fi= 66.6 N
 

Attachments

  • image.jpg
    image.jpg
    28.4 KB · Views: 469
Physics news on Phys.org
Set up equations for equilibrium. Note that there are two strings involved and thus two tensions.
 
Doc Al said:
Set up equations for equilibrium. Note that there are two strings involved and thus two tensions.

Yes, but wouldn't the tension be equally distributed among the 3 lengths of cable supporting the weight bearing pulley? That would give it an ideal mechanical advantage of 3, would it not?

Or are you saying that for mechanical advantage, I only have to take the tension of the two lengths of cable to which the force is being applied?
 
Last edited:
NanoChrisK said:
Yes, but wouldn't the tension be equally distributed among the 3 lengths of cable supporting the weight bearing pulley?
No reason to think so. Set up the equations and find out.

Assuming the usual--that the pulleys are massless and frictionless--the tension is uniform along each string. But the two strings will have different tensions.
 
Doc Al said:
No reason to think so. Set up the equations and find out.

Assuming the usual--that the pulleys are massless and frictionless--the tension is uniform along each string. But the two strings will have different tensions.

I'm just a bit lost on the setting up the equations. I know, for example, that if you just had 1 stationary pulley attached to the ceiling, you are just changing the direction of the force, so the ideal mechanical advantage is 1, giving the equation:

Ideal Mechanical Advantage = Force out / Force in = 1

And if you had a movable pulley attached to the weight, the tension would be shared by the two parts of the cable attached to the movable pulley, giving the equation:

Ideal Mechanical Advantage = 2 * Force out / force in = 2

I have a pretty good understanding of how to set up an equation for a system with just one movable pulley, or a set of pulleys attached to each other.

But I'm a bit lost when it comes to having 2 different cables and two different pulleys moving in different directions. It's not as simple as making a free body diagram as it is with just one movable pulley. How do I set up an equation for two different pulleys?
 
Last edited:
NanoChrisK said:
But I'm a bit lost when it comes to having 2 different cables and two different pulleys moving in different directions. It's not as simple as making a free body diagram as it is with just one movable pulley. How do I set up an equation for two different pulleys?
Start by labeling the tension in the two strings. Let T1 be the tension in the string that is being pulled and T2 be the tension in the other string.

Now set up an equation for each of the two movable pulleys. What forces act? What must the sum of the forces equal?
 
Doc Al said:
Start by labeling the tension in the two strings. Let T1 be the tension in the string that is being pulled and T2 be the tension in the other string.

Now set up an equation for each of the two movable pulleys. What forces act? What must the sum of the forces equal?

Ok I think I got it now. I just set up equations for both of the pulleys and substituted the T2 from the second equation for the 2T1 in the first equation. That way I could solve for T1, which also equals the force required for equilibrium.

Am I on track here?
 

Attachments

  • image.jpg
    image.jpg
    34.2 KB · Views: 451
NanoChrisK said:
Ok I think I got it now. I just set up equations for both of the pulleys and substituted the T2 from the second equation for the 2T1 in the first equation. That way I could solve for T1, which also equals the force required for equilibrium.

Am I on track here?
Exactly correct. (Except that you solved for T2 instead of T1. Fix that.)
 
Doc Al said:
Start by labeling the tension in the two strings. Let T1 be the tension in the string that is being pulled and T2 be the tension in the other string.

Now set up an equation for each of the two movable pulleys. What forces act? What must the sum of the forces equal?

Correction: substituted wrong tension. T1 = Fi, so solving for T1, Fi = 50 N
 

Attachments

  • image.jpg
    image.jpg
    37.3 KB · Views: 440
  • #10
NanoChrisK said:
Correction: substituted wrong tension. T1 = Fi, so solving for T1, Fi = 50 N
You got it. :approve:
 
  • #11
Doc Al said:
You got it. :approve:

Thanks a lot! I would not have been able to figure that out without your help!
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
24
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 29 ·
Replies
29
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K