Statics. Equilibrium With Springs and Tension Question.

In summary: Sorry, as you could see that was my first post, and I was trying to solve that question very hard.After doing it, I was so happy that I decided to share it with the others.Next times will not do that.Never mind.I have this same problem and can't figure it out for the life of me. I have been working the problem for a good 2 hours now. I got 175=2Tsin(theta)175=2T[d/(sqrt(9+d^2))]then I replaced T with kd since T=kdso 87.5=500[d^2/(sqrt(9+d
  • #1
Dieinhell100
13
0

Homework Statement


The springs BA and BC each have a stiffness of 500N/m and an unstretched length of 3m. Determine the displacement d of the cord from the wall when a force F = 175N is applied to the cord.


Homework Equations


F=ks
ƩFx=0
ƩFy=0

The Attempt at a Solution


This is as far as I got. I set out the sum of the forces in the x and y direction, assuming left is positive x and up is positive y. I don't know what to do from here with the information they have given me.

ƩFx=F-Tᴮᴬcosθ -Tᴮᶜcos(θ)=0
ƩFy=Tᴮᴬsinθ-Tᴮᶜsinθ=0
Tᴮᴬ=ks=500N/m * (l-3m)
Tᴮᶜ=ks=500N/m * (l-3m)

I'm not sure what to do when I don't have the angles or the stretched length. Attached is a drawing I made of the problem (excuse me for my messiness).
 

Attachments

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  • #2
The angles and the stretched length are related to the other side of the triangle.
 
  • #3
voko said:
The angles and the stretched length are related to the other side of the triangle.

I don't understand..?

What other side of the triangle?

And if it helps, the solution the book gives me is d=1.52m
 
  • #4
You can assume that the force of tension and the stretched lengths are equal in both springs due to symmetry. The two springs and the wall thus form an isosceles triangle. So you can relate the angle of inclination with the stretched length of the springs. That gives you another equation. You end up having three equations with three unknowns (force of tension, angle, length of the springs).
 
  • #5
Very interesting problem.
Since we have the solution, i think you should work backward to check on the given data or we need different approach such as conservation of energy or conservative force.
 
  • #6
175=2Tsin(theta)

87.5=Tsin(theta)

87.5=T[d^2/(√(3^2+d^2))]

T=ks=500*(√(3^2+d^2)-3)

d*(1-3/(√(9+d^2))=0.175

d=1.56
 
  • #7
bigu01 said:
175=2Tsin(theta)

87.5=Tsin(theta)

87.5=T[d^2/(√(3^2+d^2))]

T=ks=500*(√(3^2+d^2)-3)

d*(1-3/(√(9+d^2))=0.175

d=1.56
The protocol on this forum is to nudge in the right direction, not post complete solutions.
And you have a typo in the third line.
 
  • #8
haruspex said:
The protocol on this forum is to nudge in the right direction, not post complete solutions.
And you have a typo in the third line.


Sorry,as you could see that was my first post, and I was trying to solve that question very hard.After doing it, I was so happy that I decided to share it with the others.Next times will not do that.
 
  • #9
Never mind.
 
Last edited:
  • #10
I have this same problem and can't figure it out for the life of me. I have been working the problem for a good 2 hours now. I got
175=2Tsin(theta)
175=2T[d/(sqrt(9+d^2))]
then I replaced T with kd since T=kd
so
87.5=500[d^2/(sqrt(9+d^2)]

Can someone tell me if this is correct so far because when i simplify for d I can't seem to get the right answer.
 

1. What is statics?

Statics is a branch of mechanics that deals with the study of objects at rest or in a state of constant motion. It involves analyzing the forces acting on an object and how they affect its equilibrium.

2. How do springs affect static equilibrium?

Springs are elastic objects that can store and release energy. When a spring is in static equilibrium, it exerts a force that is proportional to its displacement from its resting position. This force can either act to restore the spring to its resting position or maintain the object's equilibrium.

3. What is tension in static equilibrium?

Tension is a force that is created by pulling on an object. In static equilibrium, tension is the force that acts on an object in opposite directions and keeps it in a state of balance.

4. How do you calculate the equilibrium point in a system with springs and tension?

The equilibrium point in a system with springs and tension can be calculated by setting the sum of all the forces acting on the object equal to zero. This includes the forces from the springs and tension, as well as any other external forces.

5. What are some real-life applications of static equilibrium with springs and tension?

Some real-life applications of static equilibrium with springs and tension include weighing scales, suspension bridges, and trampolines. These systems rely on the balance between the forces of tension and springs to maintain their stability.

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