K; the force constant of a spring

[fyzikschik]

Homework Statement

how would you find the k (N/m) of a spring that is inclined at an angle (for ex. 30degrees)? Do you just ignore the fact that it's inclided at an angle? By the way, the spring is stretched at 8cm.

Homework Equations

hooke's law: F=kx

The Attempt at a Solution

Would the forces acting on the spring in this case be force of gravity at 30deg an potential energy ( Ee=1/2 kx^2 ). I'm lost, help!

 

[billiards]

yo, i dunno. if you got hooke's law in one dimension then spose you got to just treat as a one dimensional problemo. it's just a constant of proportionality that links stress and strain.

the fact is if they were going to be nasty they'd give you some kind of anisotropy to deal with, which they haven@t (i assume) so you can deal with it as you find it!

 

[fyzikschik]

yo, i dunno. if you got hooke's law in one dimension then spose you got to just treat as a one dimensional problemo. it's just a constant of proportionality that links stress and strain.

the fact is if they were going to be nasty they'd give you some kind of anisotropy to deal with, which they haven@t (i assume) so you can deal with it as you find it!

so then i should just ignore the fact that this spring is inclined and stretched at an angle and at a a fixed distance and just say that since f=kx therefore k=mg/x ...right?

 

[radou]

so then i should just ignore the fact that this spring is inclined and stretched at an angle and at a a fixed distance and just say that since f=kx therefore k=mg/x ...right?

If there is an object of mass m attached to the spring, then the acceleration due to gravitation of the object is mg*sin(angle). By setting that up, you can solve problems 'in the direction' of the incline.

 

[Doc Al]

how would you find the k (N/m) of a spring that is inclined at an angle (for ex. 30degrees)? Do you just ignore the fact that it's inclided at an angle? By the way, the spring is stretched at 8cm.

Since the spring is stretched parallel to the incline, you need to consider the force on it parallel to the incline.

Would the forces acting on the spring in this case be force of gravity at 30deg an potential energy ( Ee=1/2 kx^2 ).

Consider the forces on the mass attached to the spring parallel to the incline. Potential energy is not a force.

so then i should just ignore the fact that this spring is inclined and stretched at an angle and at a a fixed distance and just say that since f=kx therefore k=mg/x ...right?

Wrong. The angle of the incline determines the force that stretches the spring. (Imagine if the angle of incline were 0 degrees--horizontal--or 90 degrees--vertical. Big difference, right?) Consider the forces acting on the attached mass parallel to the incline.

 

[fyzikschik]

Since the spring is stretched parallel to the incline, you need to consider the force on it parallel to the incline.

Consider the forces on the mass attached to the spring parallel to the incline. Potential energy is not a force.


Wrong. The angle of the incline determines the force that stretches the spring. (Imagine if the angle of incline were 0 degrees--horizontal--or 90 degrees--vertical. Big difference, right?) Consider the forces acting on the attached mass parallel to the incline.

ok, so the force would be gravity, k= m*sin30*9.8/x??
would force of tension be included? what about the normail force, because we are talking about the spring being in contact with the inclined launch pad

 

[fyzikschik]

I guess my real question now is, do I have to attach a mass to this spring and measure the k at the angle? Would the k be the same if I hung the spring vertically with the same mass at a 90 degree angle? Because I'm confused as to whether the k changes if the spring is at an angle or to the vertical or horizontal!

 

[Hootenanny]

I guess my real question now is, do I have to attach a mass to this spring and measure the k at the angle? Would the k be the same if I hung the spring vertically with the same mass at a 90 degree angle? Because I'm confused as to whether the k changes if the spring is at an angle or to the vertical or horizontal!

Your above answer is correct. The normal reaction force does not affect the tension in the spring since it acts perpendicular to the spring. With respect to your next question, not the spring constant remains constant. Although by altering the angle you are varying the force applied to the spring, the spring will extend by an amount equal to x = F/k. This is why it is called the spring constant. It is constant. To verify this, you could try varying the angle of the incline and noting the extension produced in the spring.

 

[fyzikschik]

so then, at an angle theta, k=*m*sintheta*9.8/x ??
i'm assuming we can discard the force of tension and the normal force (since they are perpendicular)

 

[Doc Al]

ok, so the force would be gravity, k= m*sin30*9.8/x??
would force of tension be included?

The forces acting on the mass parallel to the incline are:
(1) the parallel component of gravity, which is mg sin30
(2) the spring force, which is kx.
Set those equal--since the mass is in equilibrium-- and you'll get k as you described.

what about the normail force, because we are talking about the spring being in contact with the inclined launch pad

The normal force will balance the perpendicular component of gravity. But that won't help you find k, so you don't need to worry about it.

I guess my real question now is, do I have to attach a mass to this spring and measure the k at the angle? Would the k be the same if I hung the spring vertically with the same mass at a 90 degree angle? Because I'm confused as to whether the k changes if the spring is at an angle or to the vertical or horizontal!

k is a property of the spring. I doesn't matter what the angle of incline is, the spring constant is the same. If you change the angle, both the sine of the angle and the amount of stretch will change together. The value of k that you find will be the same no matter what the angle. So if you've measured it once, that's enough.

 

[Hootenanny]

so then, at an angle theta, k=*m*sintheta*9.8/x ??
i'm assuming we can discard the force of tension and the normal force (since they are perpendicular)

Do you mean the tension in the spring. And yes you can ignore the normal force.

 

[fyzikschik]

i'm going to be given a launch angle and a distance and i have to determine how far i need to pull this spring back to hit a target. I'm assuiming (and according to doc al) that since the force constant is the same that i should just measure the mass of the spring, and say (k=m*g*x). I've left out the angle because from what you said, i understand that the angle does not affect k, am I correct in thinking so?

 

[fyzikschik]

and to further clarify, no mass will be attached to the spring. only the spring itself will be in motion.

 

[Doc Al]

i'm assuiming (and according to doc al) that since the force constant is the same that i should just measure the mass of the spring, and say (k=m*g*x).

I never said anything about the mass of the spring. I thought you were asking about how to measure the spring constant of a (presumably massless) spring by hanging a mass off one end. If the spring constant is high enough (in other words, if the spring is stiff enough) you can neglect the mass of the spring in measuring its spring constant.

Realize that the mass of the spring is distributed over the length of the spring, so it is not the same as simply hanging a mass off one end.

(It pays to specify the full situation as exactly as possible so we can understand the context of your question.)

 

[fyzikschik]

thanks for being patient, doc.
all I'm trying to do is figure out if the spring constant of a spring can be determined by its own mass and its own length. and if that k would apply if i was to use it in a calculatation once the spring is on an inclined plane.

the main purpose of this is to be able to launch the spring (manually using a pen to extend it at a calculated extension x) so that it can hit a target. this is why i need to know if k is the same at an inclined angle as it is if i were just to say k=mg/x.

we are given a launch angle on test day and are given a distance (of target). i don't know the launch angle, so i have to figure out a k that would apply for everything. and this is where I'm lost. because i don't know if the components of gravity will affect the k once it is on the inclined angle.

i'm new to physics, and I'm graduating and i would be grateful if u could continue to help.

if i hung the spring vertically, and attach a mass to it, and find k that way, would that k apply to calculations such as:

x= square root (mv^2/k) <--this determines how far i need to pull the spring back at the angle to get it to land on the target. v is found through ( v= square root of gx/sin2theta)

how would u find k in such a situation?

 

[Doc Al]

My advice: Keep it simple. Unless that spring is a whopper, I would ignore its mass in measuring the spring constant. Hang it unloaded, measure its unstretched position. Then hang a mass on it and measure its stretch. Use that data to measure k.