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Calculating moments with springs/Hooke's Law

Hi guys,

I am a little confused on where to begin with this question. It involves moments and Hooke's law. I have tried to take moments about the point X, but end up with an algebraic distance,
p, of .

I need a few hints on this question. How should I begin to approach it?


"A uniform rod AB of weight W, with a length of 20cm, is suspended by 2 vertical springs X and Y attached to the rod at A and B respectively. The upper ends of the springs are attached to a horizontal beam. When the springs are unextended, they have the same length. The tension TX in X is given by TX = kx and the tension TY in Y is given by TY = 3ky, where k is a constant and x and y are extensions of X and Y.

At what distance from A must a body of weight 5W be attached to the rod of the rod is to be horizontal?"
Original post by theoinkk
Hi guys,

I am a little confused on where to begin with this question. It involves moments and Hooke's law. I have tried to take moments about the point X, but end up with an algebraic distance,
p, of .

I need a few hints on this question. How should I begin to approach it?


Have you drawn a diagram showing the forces acting?

My first impression is to set up 2 equations.
The vertical forces are balanced.
The downwards force total is W+5W = 6W
If the rod is horizontal then the two extensions are equal.
This means the tension at end Y is 3 times that at end X
Let tension at X be T
So upward forces are T + 3T

This gves me
6W = 4T

Now take moments about one of the ends and eliminate either T or W form the equation.

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