Parallel Springs subject to load

AI Thread Summary
The discussion focuses on determining the equivalent spring constant (ke) for a system of two parallel springs attached to a massless rigid bar under an applied force. The approach involves equating the potential energy of the two springs to that of the equivalent spring, using small angle approximations for angular displacements. Participants noted the need for a geometric relationship between the displacements of the springs and the total spring compression. The equilibrium conditions require that the sum of vertical forces and torques about the point of applied force must equal zero. Ultimately, the goal is to derive an expression for ke based on these relationships.
pige
Messages
3
Reaction score
0

Homework Statement


A system of two springs is considered, with spring constants k1 and k2, arranged in parallel and attached to a massless rigid bar. The bar remains horizontal when no force on it is zero. Determine the equivalent spring constant (ke) that relates the force applied a distance l1 from one spring and l2 from the other (F is between the springs) to the resulting displacement for small oscillations


Homework Equations


F1=k1*x1
F2=k2*x2

F=ke*x
sin(theta)=theta
U=1/2*k*x^2

T=F*R

The Attempt at a Solution



It was decided after much group work to try and use energy. We ended up setting the potential energy of the two springs to the potential energy of the equivalent spring. Using the small angle approximation we could write the angular displacements as the displacement of the springs due to the rotation, but weren't able to derive an expression for the translational displacement of the bar, and thus not able to get an expression for the total spring compression.
 
Physics news on Phys.org
A picture would help.
 
Spinnor said:
A picture would help.

13z32wx.jpg
 
The force F causes the two springs to compress until equilibrium is reached. At that point the sum of the forces in the vertical direction sum to zero. Also the sum of the torques about the point the force F is applied is zero.
 
Spinnor said:
The force F causes the two springs to compress until equilibrium is reached. At that point the sum of the forces in the vertical direction sum to zero. Also the sum of the torques about the point the force F is applied is zero.

That may be true but it is not enough.

You get three relationships, sum forces in vertical = 0, sum of torques about the point of applied force F = 0, and a geometrical relationship between x1, x2, and x,3. See below. Solve for ke.
 

Attachments

  • springs241.jpg
    springs241.jpg
    13.8 KB · Views: 2,404
  • springs 2242.jpg
    springs 2242.jpg
    14 KB · Views: 1,842
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top