- #1
GeoMike
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Ok, this seems simple, and it probably is, but I'm just not seeing why I'm getting two non-equivalent solutions...
"A massless platform is supported by 4 equal springs (same length, same spring constant, etc.)
A mass m is placed on the center of the platform causing an equal compression of x in each spring. What is the spring constant k of a single spring in terms of the mass m, the acceleration of gravity g, and the compression of the spring x?
Fspring = -kx
Fgrav = mg
Uspring = 1/2(kx^2)
Ugrav = mgx
What's throwing me off is that there seem to be two valid ways to solve this, each resulting in a different value of k...
If I use the fact that the force acting on each spring is 1/4 the force of gravity acting on the mass I get:
kx=mg/4
So... k = (mg)/(4x)
But, then isn't the elastic potential energy stored in each spring 1/4 the gravitational potential energy in the mass when it is first placed on the platform (before any compression)? If so, I get:
1/2(kx^2) = mgx/4
So... k = (mg)/(2x)
The back of the book gives (mg)/(4x). Which I understand, the problem is that the second answer seems valid to me, but obviously the second one is wrong. I just can't see where is my error in reasoning. Help?
Thank you!
-GM-
Homework Statement
"A massless platform is supported by 4 equal springs (same length, same spring constant, etc.)
A mass m is placed on the center of the platform causing an equal compression of x in each spring. What is the spring constant k of a single spring in terms of the mass m, the acceleration of gravity g, and the compression of the spring x?
Homework Equations
Fspring = -kx
Fgrav = mg
Uspring = 1/2(kx^2)
Ugrav = mgx
The Attempt at a Solution
What's throwing me off is that there seem to be two valid ways to solve this, each resulting in a different value of k...
If I use the fact that the force acting on each spring is 1/4 the force of gravity acting on the mass I get:
kx=mg/4
So... k = (mg)/(4x)
But, then isn't the elastic potential energy stored in each spring 1/4 the gravitational potential energy in the mass when it is first placed on the platform (before any compression)? If so, I get:
1/2(kx^2) = mgx/4
So... k = (mg)/(2x)
The back of the book gives (mg)/(4x). Which I understand, the problem is that the second answer seems valid to me, but obviously the second one is wrong. I just can't see where is my error in reasoning. Help?
Thank you!
-GM-