Quick Static Equilibrium Questions

In summary, the force exerted on the beam by the hinge has a vertical component that is nonzero and up. The magnitude T of the tension force of the upper cable is 16 N. Using the equations for torques and tension, the correct logic is to work through each pulley and solve for the tension in the top string, which is found to be 8P, or 16 N. This method takes into account the different tensions in each rope and leads to the correct answer.
  • #1
pietastesgood
30
0

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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  • #2
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.)
 
  • #3
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?
 
  • #4
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)
 
  • #5
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.
 
  • #6
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?
 
  • #7
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.
 
  • #8
Awesome! Thanks, both of you!
 

1. What is static equilibrium?

Static equilibrium is a state in which all forces acting on an object are balanced, resulting in no net acceleration or movement. In other words, the object is at rest or moving at a constant velocity.

2. How do you determine if an object is in static equilibrium?

An object is in static equilibrium if the sum of all forces acting on it is equal to zero and the sum of all torques (rotational forces) acting on it is also equal to zero. This can be determined by using the equations of static equilibrium, which include the equations for force balance and torque balance.

3. What are some examples of objects in static equilibrium?

Some examples of objects in static equilibrium include a book sitting on a table, a ladder leaning against a wall, and a person standing still on the ground. In each of these cases, the forces acting on the object are balanced, resulting in no net movement.

4. How does the center of mass affect static equilibrium?

The center of mass is an important factor in determining static equilibrium because it is the point at which an object's mass is evenly distributed. If an object's center of mass is directly above its base of support, it will be in stable equilibrium, meaning it will not tip over or move. If the center of mass is outside of the base of support, the object will not be in static equilibrium and will fall or topple over.

5. What are some real-world applications of static equilibrium?

Static equilibrium is important in many engineering and architectural designs, such as bridges, buildings, and structures. It is also relevant in sports and recreation, such as balancing on a surfboard or bicycle. Additionally, understanding static equilibrium is crucial in the field of biomechanics for studying the balance and stability of human movement.

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