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

In summary, to lift a 200 N load using an arrangement with two movable pulleys, two different tensions must be considered. Using the equations for equilibrium and the fact that the pulleys are massless and frictionless, the tension in the pulleys can be calculated and used to find the input forces needed for lifting the load.
  • #1
NanoChrisK
14
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
 

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  • #2
Set up equations for equilibrium. Note that there are two strings involved and thus two tensions.
 
  • #3
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?
 
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  • #4
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.
 
  • #5
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?
 
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  • #6
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?
 
  • #7
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?
 

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  • #8
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.)
 
  • #9
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
 

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  • #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!
 

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

What is a tricky pulley problem?

A tricky pulley problem is a type of physics problem that involves a system of pulleys and ropes. These problems often require the use of Newton's laws of motion and the concept of mechanical advantage to find the solution.

How do you approach solving a tricky pulley problem?

The first step in solving a tricky pulley problem is to draw a clear diagram of the system, including all of the pulleys and ropes. Then, identify all of the forces acting on each pulley, and use Newton's laws of motion to set up equations. Finally, use the concept of mechanical advantage to simplify the equations and find the solution.

What is the concept of mechanical advantage in a pulley system?

Mechanical advantage is the ratio of the output force to the input force in a system. In a pulley system, the mechanical advantage is equal to the number of ropes supporting the load. For example, a system with two ropes supporting the load has a mechanical advantage of 2.

What are some common mistakes when solving tricky pulley problems?

One common mistake is not including all of the forces acting on each pulley in the equations. Another mistake is not considering the direction of the forces, which can result in incorrect solutions. It is also important to double-check all calculations and units to avoid errors.

Are there any tips or tricks for solving tricky pulley problems?

One helpful tip is to start by assuming that the ropes and pulleys are massless and frictionless, and then add in any relevant factors later. It can also be helpful to label the forces on the diagram and use a consistent direction for positive and negative forces. Additionally, practicing with different types of pulley problems can help develop problem-solving strategies.

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