How Can You Find Forces P, F, and T in a Static Equilibrium Problem?

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SUMMARY

This discussion focuses on solving for the forces P, F, and T in a static equilibrium problem involving inextensible rods. The key equations used are ΣFx = 0 and ΣFy = 0, which indicate that the sum of forces in both the x and y directions must equal zero. Participants emphasize the importance of considering internal forces and the geometry of the system, particularly using trigonometric relationships to express forces in terms of angles. The final expressions for forces P and F involve the sine and cosine of specific angles derived from the geometry of the setup.

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Handsome jack
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1. Homework Statement

Find forces P,F,and T

Homework Equations


Fx = 0
Fy = 0

The Attempt at a Solution


b2tdBTh.jpg

So far I only got force T from getting the moment from point b. Don't know how-to start getting force P and F since they're both at the same point. Any help will be much obliged thank you
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Assuming the rods are inextensible, find the reaction between the velocities of the three ends.
For example, if a horozontal rod is given a velocity of say 5m/s at one end A along the direction of rod, then the other end will also have the same velocity. If it does not, the rod will extend.
 
Welcome to PF!

What do you mean by Fx = 0 and Fy = 0? Have you tried to use these?
 
TSny said:
Welcome to PF!

What do you mean by Fx = 0 and Fy = 0? Have you tried to use these?
That means that all forces on x and y-axis must equal to 0 since it's on equilibrium
 
Handsome jack said:
That means that all forces on x and y-axis must equal to 0 since it's on equilibrium
Do you mean that the sum of the x components of all of the forces must be zero? Similarly for the y components?

If so, have you tried using these facts?
 
How can the system be at equilibrium? There is a net force acting along horizontal.
 
Titan97 said:
How can the system be at equilibrium? There is a net force acting along horizontal.
Looks like the force P might be acting down and to the left according to the second picture in post #1.
 
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If its not at equilibrium, then can't I say that the force along the rod at both ends should be equal? (Assuming the rod does not extend)
 
Oh God sorry. Force P is down to the left. It's on equilibrium, just have to find what the other forces are that are F and p that contribute to the cancellation of the other forces
 
  • #10
Titan97 said:
If its not at equilibrium, then can't I say that the force along the rod at both ends should be equal? (Assuming the rod does not extend)
No, I don't think so. A rod can sustain tension. There is no need for the force parallel to the rod at one end to equal the force parallel to the rod at the other end.

Also, if you were to consider a single rod in the system, you would need to take into account forces acting on the rod from the other rods. If the rod is in equilibrium then the net force (including internal forces) acting parallel to the rod at one end would be equal and opposite to the net parallel force at the other end (assuming we can neglect the weights of the rods.)
 
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  • #11
Handsome jack said:
Force P is down to the left. It's on equilibrium, just have to find what the other forces are that are F and p that contribute to the cancellation of the other forces
How would you write out ΣFx = 0 for this problem?
 
  • #12
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tried using process of elimination to find P and F so far I got the answer not sure if it's correct
 
  • #13
OK, your process looks correct. I'm too lazy today to get out my calculator and check all the numbers.
 
  • #14
TSny said:
OK, your process looks correct. I'm too lazy today to get out my calculator and check all the numbers.
Thanks
 
  • #15
I put the equations into some software. It gives somewhat different answers than what you are getting. Could be round off errors. For example sin(33.69) = .5547 instead of .55.

Note that using the little triangles allows you to have exact expressions for the sines and cosines of the angles. For force P, the hypotenuse of the triangle is √10 so you can write the sine of the angle as 1/√10 and the cosine as 3/√10.

Force P then comes out to be 2400√10 . Likewise F can be expressed as a nice number times √13.
 
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