# AS mechanics springs physics

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First of all the forces going up must equal the forces going down since the object is stationary.

Secondly, place an imaginary pivot at the centre of mass. Now take the moments about the pivot. There must be a balance here otherwise the sign would rotate.

You will end up with two simultaneous equations and the unknowns are the extensions in both springs. I've done the working out and got an answer of 0.0609m.

Note, the springs impart a force on the sign which in this case, acts "upwards" and you use Hooke's law where F = kx. So the sign imparts a force on the spring, causing it to extend downwards and the spring is pulling the sign upwards (Newtons Third law pair).

Last edited by 0le; 3 weeks ago

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(Original post by

It is a system of equations involving balance of forces and balance of moments.

First of all the forces going up must equal the forces going down since the object is stationary.

Secondly, place an imaginary pivot at the centre of mass. Now take the moments about the pivot. There must be a balance here otherwise the sign would rotate.

You will end up with two simultaneous equations and the unknowns are the extensions in both springs. I've done the working out and got an answer of 0.0609m.

Note, the springs impart a force on the sign which in this case, acts "upwards" and you use Hooke's law where F = kx. So the sign imparts a force on the spring, causing it to extend downwards and the spring is pulling the sign upwards (Newtons Third law pair).

**0le**)It is a system of equations involving balance of forces and balance of moments.

First of all the forces going up must equal the forces going down since the object is stationary.

Secondly, place an imaginary pivot at the centre of mass. Now take the moments about the pivot. There must be a balance here otherwise the sign would rotate.

You will end up with two simultaneous equations and the unknowns are the extensions in both springs. I've done the working out and got an answer of 0.0609m.

Note, the springs impart a force on the sign which in this case, acts "upwards" and you use Hooke's law where F = kx. So the sign imparts a force on the spring, causing it to extend downwards and the spring is pulling the sign upwards (Newtons Third law pair).

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(Original post by

I still dont understand, could you show me the working out you did in the simultaneous equation please?

**gorilla_08**)I still dont understand, could you show me the working out you did in the simultaneous equation please?

The first equation is a balance of forces in the vertical direction. You have one force going down and two going up (the restoring forces of each spring). The two upwards forces contain two unknowns which will be the extensions in the spring, x1 and x2. Use the general spring force equation, F = kx where F is the restoring force in the spring, k is the spring constant and x is the extension.

So two spring forces up = one gravity force down.

**This is equation one.**

You use the same forces to do a moments equation about the pivot. You ignore gravity for this second equation because it acts at the pivot and therefore the moment arm, the distance in this case, is zero.

One spring force to the left of the pivot multiplied by distance = One spring force to the right of the pivot multiplied by distance.

**This is equation two.**

You have two simultaneous equations with two unknowns, x1 and x2, which are the extensions in each spring.

Please show us your working out.

Last edited by 0le; 2 weeks ago

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