Calculating force from a system of connected springs

Click For Summary
SUMMARY

The discussion focuses on calculating the total force exerted on two identical masses connected by springs in a frictionless environment. The force exerted by each spring is defined by Hooke's Law, F = -kx. The correct total force on each mass is determined to be F(total) = (k1 + 2k2)(x), where k1 is the force constant of the wall springs and k2 is the force constant of the connecting spring. The confusion arises from the misunderstanding that the force from the middle spring is not divided but rather doubled due to its total stretch of 2x.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Knowledge of spring constants (k1 and k2)
  • Concept of equilibrium in mechanical systems
  • Basic principles of forces in a frictionless environment
NEXT STEPS
  • Study the implications of multiple springs in series and parallel configurations
  • Learn about the dynamics of oscillating systems involving springs
  • Explore advanced applications of Hooke's Law in real-world scenarios
  • Investigate the effects of damping on spring systems
USEFUL FOR

Students of physics, mechanical engineers, and anyone interested in understanding the dynamics of spring systems and force calculations in equilibrium conditions.

miaou5
Messages
13
Reaction score
0

Homework Statement



http://img854.imageshack.us/img854/1828/physicssprings.png

Two identical masses on a frictionless surface are attached to two walls by springs, each with a force constant k1, and to each other by a spring with force constant k2. (See figure, where system is at equilibrium) For simplicity, assume the springs are massless. If the left mass is moved a distance x to the left and the right mass is moved a distance x to the right, the magnitude of the force on each mass is equal to what in terms of k and x (where x signifies displacement)?

Homework Equations



The general equation for force resulting from a spring is F = -kx, where k is the force constant and x is the displacement.

The Attempt at a Solution



I'm taking the mass on the right. Assume it is moved a distance x to the right; this would mean that the magnitude of force due to spring 1 (the spring connected the wall) would be F1 = (k1)(x). (The direction of the force would be to the left.)

The middle spring (spring 2) is stretched a total 2x (x from each mass); from what I learned, however, since it is stretched x to the right and x to left, this means that there would be two resultant forces--a force (k2)(x) going to left (exerted on the mass to the right), and a force (k2)(x) going to the right (exerted on the mass to the left).

Since the question asks for the force exerted on EACH mass, I'll take the mass on the right:

F(total) = F1 (force exerted on mass by spring 1) + F2 (force exerted on mass by spring 2) = (k1)(x) + (k2)(x) = (k1 + k2)(x)

However, the correct answer is (k1 + 2k2)(x). How is this possible? This would mean that spring 2 exerts a force of 2(k2)(x) to the left on the right mass, but it should only be exerting a leftwards force of (k2)(x). Could somebody please explain this? Thank you so much in advance!
 
Last edited by a moderator:
Physics news on Phys.org
If a spring is stretched by amount x, the force it exerts on each end is kx. The force isn't 'divided' between the two ends. (Note that you didn't divide the force from the rightmost spring in two!)

Since the middle spring is stretched by 2x, the force it exerts will be k(2x).
 
Thank you so much Doc Al! [Ah, finally the problem makes sense!]
 

Similar threads

Replies
17
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
Replies
6
Views
1K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
29
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K