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Hookes law

 
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Sep8-11, 10:13 PM   #1
 

Hookes law


1. The problem statement, all variables and given/known data
Whether or not the spring is vertical or not, assuming that the spring is ideal, is Fnet always equal to -kx?

Is it incorrect to write the sum of the forces of a vertical spring as Fnet = -kx - mg?

And does hooke's law account for the force of friction on a horizontal spring? If not would the sum of the forces look like this? Fnet = -kx - [itex]\mu[/itex]N


2. Relevant equations
F = -kx


3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 
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Sep8-11, 10:25 PM   #2
 
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Quote by joej24 View Post
1. The problem statement, all variables and given/known data
Whether or not the spring is vertical or not, assuming that the spring is ideal, is Fnet always equal to -kx?
No, for a vertical spring, you do have to include gravity.

Quote by joej24 View Post
Is it incorrect to write the sum of the forces of a vertical spring as Fnet = -kx - mg?
For a vertical spring, this is in fact correct.

Quote by joej24 View Post
And does hooke's law account for the force of friction on a horizontal spring? If not would the sum of the forces look like this? Fnet = -kx - [itex]\mu[/itex]N
Hooke's law doesn't include friction automatically. Hooke's law has to do with the spring restoring force only. If there is friction, you have to include it explicitly. The way you did it there is ok, but be careful with the signs. Friction always opposes the motion, so if the mass is moving in the + direction, the frictional force is negative, and if the mass is moving in the - direction, the frictional force is positive. So there isn't one equation that will be correct at all times.
 
Sep8-11, 10:41 PM   #3
 
If a vertical spring is hanging and at rest, its net force = 0.
So that means F = 0 = -kx - mg, but doesn't the force of the spring oppose the force of gravity? How does this make sense? The force of gravity always points downward ( is negative), but -kx is negative when the spring is stretched and at rest.
 
Sep10-11, 02:31 PM   #4
 
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Hookes law


Quote by joej24 View Post
If a vertical spring is hanging and at rest, its net force = 0.
So that means F = 0 = -kx - mg, but doesn't the force of the spring oppose the force of gravity? How does this make sense? The force of gravity always points downward ( is negative), but -kx is negative when the spring is stretched and at rest.
This is a case where you have to be careful about signs. First of all, the length that goes into Hooke's law is a displacement. I'll use the symbol 'y' instead of 'x', since y is usually used for the vertical position coordinate, and x for the horizontal position coordinate. Hooke's law should really be written as:

F = -kΔy

where

Δy ≡ y - y0

In this expression, y0 is the position at which the spring is neither stretched nor compressed. The reason we can write F = -ky is because, in general we can choose the origin of our coordinate system to be at y0 (i.e. we can choose y0 = 0) so that all other positions are measured from this reference point. But the key point here is that y is the spring displacement and as such, it is a vector. At the very least, for a 1D situation, y must have an intrinsic sign (+ or -) that tells you which direction this displacement is in.

Next, we come to sign conventions. This is another case where we have the freedom to choose something. Gravity points downward. So, if you write F = -mg, you've implicity chosen the sign convention that "downward is negative" (i.e. downward is the negative y direction). If so, this affects the sign of the displacement y. In the force balance equation,

F = -mg - ky,

if you stretch the spring, then the position of the mass goes down below the position at which the spring is undisplaced. Since downward is negative, this means that the displacement is negative: y < 0. But if y is negative, then it follows that the spring force is positive: -ky > 0. So, when the spring is stretched, the spring force is positive and points upward, counteracting gravity, just as you would expect. Similarly, if the spring is compressed, then the position, y, of the mass goes up above the position at which the spring is undisplaced. In other words the displacement is positive: y > 0. As a result, -ky < 0 and a compressed spring has a restoring force that points downward, just as you would expect.

You are free, of course, to choose the opposite sign convention in which "downward is positive". Under this sign convention, we write gravity as F = +mg, and the total force becomes:

F = mg - ky

Now, if the mass is initially at y = 0 (the position for which the spring is undisplaced) and then you stretch it by a small amount, its new position is physically below the zero-point. Since downward has been defined to be positive, it follows that the displacement is positive: y > 0. Therefore -ky < 0, and the spring force points in the negative direction (upward) just as one would expect. If the spring is compressed, the the mass moves above the zero-point. Since upward is negative, it follows that the displacement is negative: y < 0. As a result, -ky > 0, and the force for a compressed spring points in the positive (downward) direction as one would expect.

This second sign convention is perhaps more intuitive, since a positive displacement in y corresponds to spring extension (lengthening) and a negative displacement in y corresponds to spring compression (shortening).
 
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