Spring problem with G.E

Apr 2010
30
0
Question

Consider a system of 2 masses on a line connected by three springs with spring constants \(\displaystyle k1, k2\) and \(\displaystyle k3\). The system looks like

|- - -m - - -m- - - -|

where m indicates a mass, dash lines indicate a spring and the vertical lines indicate a wall. Take the origin of coordinates to be the first wall on the left. Let \(\displaystyle x1\) denote the position of the first mass and \(\displaystyle x2\) denote the position of the second mass. The distance between the walls is \(\displaystyle L\). The problem is to compute equilibrium position for the masses.

a) Write down the force on each mass.

b) Set the forces equal to zero

c) Solve the system of equations with Gaussian Elimination.

Answer

i arrived at this equation and i have no idea if i am on the right track

\(\displaystyle x2(-k3-k2/k2 + {k2/k2+k3})= -k3L/k2\)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Question

Consider a system of 2 masses on a line connected by three springs with spring constants \(\displaystyle k1, k2\) and \(\displaystyle k3\). The system looks like

|- - -m - - -m- - - -|

where m indicates a mass, dash lines indicate a spring and the vertical lines indicate a wall. Take the origin of coordinates to be the first wall on the left. Let \(\displaystyle x1\) denote the position of the first mass and \(\displaystyle x2\) denote the position of the second mass. The distance between the walls is \(\displaystyle L\). The problem is to compute equilibrium position for the masses.

a) Write down the force on each mass.

b) Set the forces equal to zero

c) Solve the system of equations with Gaussian Elimination.

Answer

i arrived at this equation and i have no idea if i am on the right track

\(\displaystyle x2(-k3-k2/k2 + {k2/k2+k3})= -k3L/k2\)
You "arrived at this equation" to answer what question? (a) asked for "the force on each mass" which asks for two force values, not one equation.

The first mass, at x1, has a force due to the spring on its left equal to -k1x1 (negative since it is to the left) and on the right equal to k2(x2- x1) since the length of the spring on the right is x2-x1. The total force on the first mass is k2(x2- x1)- k2x1.

The force on the second mass will be the force due to the middle spring, which is opposite but the same strength as its force on the first mass, -k2(x2- x1). The length of the third spring is L- x2.
 
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Apr 2010
30
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oh right, i never considered the system using hooke's law.

and the force on the 3rd spring would be k3(L-x2)?