Quick Static Equilibrium Questions

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Homework Help Overview

The discussion revolves around static equilibrium problems involving a horizontal beam supported by a hinge and cable, as well as a system of pulleys and weights. Participants are exploring the forces acting on the beam and the tension in the cables, particularly focusing on the relationships between these forces and the setup of free-body diagrams.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the necessity of a vertical component of force from the hinge, considering the tension in the cable. There is also confusion regarding the setup of free-body diagrams for the pulley system and the effects of multiple pulleys on the required pulling force.

Discussion Status

Some participants have offered insights into the relationships between tensions in the pulleys and the forces acting on the block. There is ongoing exploration of the assumptions regarding tension consistency across the pulleys, with various interpretations being discussed without reaching a definitive consensus.

Contextual Notes

Participants are navigating the complexities of pulley systems and static equilibrium, with specific attention to the effects of frictionless pulleys and the distribution of tension. There is mention of homework constraints that may limit the information available for analysis.

pietastesgood
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Homework Statement



A horizontal beam of weight W is supported by a hinge and cable as shown. The force exerted
on the beam by the hinge has a vertical component that must be:
2yy4jsg.jpg


Answer is nonzero and up.

The pull P is just sufficient to keep the 14-N block and the weightless pulleys in equilibrium
as shown. The magnitude T of the tension force of the upper cable is:
v475kw.jpg


Answer is 16 N.

Homework Equations



1. τ=r x F
W=mg
F=ma
Tsin(θ)=mg

2. T=mg+P

The Attempt at a Solution



1. Not quite sure why the answer for this one is nonzero and up. Shouldn't the cable already have a tension force in the y direction that cancels out the weight of the beam? Then the only force that the hinge needs to exert is in the positive x direction to counteract the tension in the y-direction. Can someone shed some light on this one?

2. I'm having a bit of trouble setting up a free-body diagram for this one, but I do know that pulleys reduce the required pulling force by a factor of 2, and since there are three pulleys it reduces the pulling force by a factor of 8. 14/8 is 1.75, which using T=mg+P equals 15.75N for T, or 16 N. Is this logic even correct? And if so, could someone help me set up the FBD so I can prove my logic?
 
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pietastesgood said:
1. Not quite sure why the answer for this one is nonzero and up. Shouldn't the cable already have a tension force in the y direction that cancels out the weight of the beam? Then the only force that the hinge needs to exert is in the positive x direction to counteract the tension in the y-direction. Can someone shed some light on this one?
Consider torques about the point where the cable attaches to the beam.

2. I'm having a bit of trouble setting up a free-body diagram for this one, but I do know that pulleys reduce the required pulling force by a factor of 2, and since there are three pulleys it reduces the pulling force by a factor of 8. 14/8 is 1.75, which using T=mg+P equals 15.75N for T, or 16 N. Is this logic even correct? And if so, could someone help me set up the FBD so I can prove my logic?
Just work your way through each pulley, starting with the one at the bottom. If the tension in that pulley's rope is P, what must be the tension in the rope above it? And so on. Then you can solve for P and use it to answer the question. (Draw free body diagrams for each pulley and for the block.)
 
Oh, understand the first one now, thanks!

10oog8i.jpg

I'm not sure if I got it. P+42/3=14N, thus P=2. T=14+P=16N, which is right, but then the top string has two different tensions. What did I do wrong?
 
I think as the pulleys are considered friction-less the tension in the string over it will remains same everywhere. If it is P on the right side part of the string it will be P on the left side as well. If we consider equilibrium of lowest mass less pulley, what should be the tension in the string over the middle pulley? (tensions in the three strings will not be same)
 
pietastesgood said:
I'm not sure if I got it. P+42/3=14N, thus P=2. T=14+P=16N, which is right, but then the top string has two different tensions. What did I do wrong?
For one thing, you seem to have assumed that the three ropes attached to the block have the same tension. Do not assume any such thing.

Start with the bottom pulley. Since the tension is P, the rightmost rope attached to the block must have tension P. Then keep going until you've analyzed all the pulleys and the block.
 
Hm, well if that's the case, then the tension in the string over the middle pulley is 2P, and the tension in the string over the top pulley is 4P. So together the tensions equal 4P+2P+P=7P, which has to balance out the 14N weight. So P=2. And since the tension in the top string equals 14+P, the tension is 16N?
 
pietastesgood said:
Hm, well if that's the case, then the tension in the string over the middle pulley is 2P, and the tension in the string over the top pulley is 4P. So together the tensions equal 4P+2P+P=7P, which has to balance out the 14N weight. So P=2.
You got it! :approve:

And since the tension in the top string equals 14+P, the tension is 16N?
I would say that the tension in the top string must equal 8P = 16N, since the string through the top pulley has a tension of 4P.
 
Awesome! Thanks, both of you!
 

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