# Parallel springs

## Homework Statement

In the parallel spring system, the springs are positioned so that the 30 N weight stretches each spring equally. The spring constant for the left hand spring is 3.2 N/cm and the spring constant for the right-hand spring is 6.6 N/cm.
The diagram is basically 2 springs holding a 30 N weight. (similar to this picture)http://img53.imageshack.us/my.php?image=simg476yr9.gif

The question is asking how far down will the 30N weight stretch the springs?

## The Attempt at a Solution

30 N/ (3.2 N/cm) = 9.375 cm
30 N / (6.6 N/cm) = 4.54 cm

I'm not entirely sure what to do from this point because conceptually, I don't understand how one spring with a smaller spring constant can hold up a weight at the same level as the larger spring constant. If you know how to do this problem, please help! This problem is due by Monday! Thank you in advance to those who reply!!

Related Introductory Physics Homework Help News on Phys.org
u can't divide 30 by the spring constants because it is not the force applied to each,
for equal strectching, the forces on the springs have to be different.
how can u make the force different on two points by using one object?
try using torque idea (in this case moment is enough too)

I'm sorry. Perhaps you could clarify what you meant by torque. I haven't learned this concept in class yet and I tried looking it over the internet but I'm not sure how to apply the equation i found for torque (torque = r X F )

Doc Al
Mentor
I'm not entirely sure what to do from this point because conceptually, I don't understand how one spring with a smaller spring constant can hold up a weight at the same level as the larger spring constant.
Note that it tells you that both springs are stretched equally. Call the unknown stretch "X". What force does each spring exert? What must the total force from both springs equal?

ahhh...
3.2X + 6.6X =30

therefore X = 3.06 cm

Is this the right way to go about it?

Doc Al
Mentor
You got it.