Question related to SHM and free body diagrams

[ehabmozart]

Homework Statement

Draw three free-body diagrams showing the forces acting on a toy suspended by a vertical spring. The diagrams should show the situations when the toy is at its equilibrium position, and above and below the equilibrium position.

Homework Equations

The Attempt at a Solution

At eq. it is fine.. Mg=F restoring (though i don't believe that restoring force acts in the eq. postion cs F=-kx)... above eqm, mg + air resistance > k(x - d) ... (how is this possible).. when it is above F restoring should be inline with mg... my question is what gives the toy its upward motion... ; below eqm, mg < air resistance + k(x + d). (again this time mg dominates) the reason it actually moves down... These are the answers given by my textbook... i feel that they r the other way round.. I need help in understanding the concept URGENTLY and thanks to whoever gives me a GOOD REPLY !

 

[Doc Al]

Forget about air resistance for the moment. Just compare the weight (mg) with the restoring force. Note that the restoring force depends on the amount of stretch in the spring--measured from its unstretched length.

 

[ehabmozart]

Let us take above equillibrium position.. Where's mg pointing.. Always downwards.. What bout F.. again downwards... No what gives the upward motion.. I am confused...

 

[Doc Al]

Let us take above equillibrium position.. Where's mg pointing.. Always downwards.. What bout F.. again downwards... No what gives the upward motion.. I am confused...

Why do you think the restoring force acts downwards?

 

[ehabmozart]

Restoring force acts towards the equillibrium position.. Isn;t it?

 

[Doc Al]

Restoring force acts towards the equillibrium position.. Isn;t it?

No, the force of the spring always acts to pull back towards the unstretched position. In the equilibrium position (once there's a mass hanging from the spring) the spring is stretched and thus is exerting a force.

(Note: The net force on the mass--the sum of the weight plus spring force--is towards the equilibrium position.)

 

[ehabmozart]

Pardon my dumbness! ... U said the force pulls back the unstretched spring towards equillibrium... So my words were right!

 

[Doc Al]

Pardon my dumbness! ... U said the force pulls back the unstretched spring towards equillibrium... So my words were right!

The net force, not the spring force. But you need to understand where the net force comes from and why it always acts towards the equilibrium point (which is not the unstretched position).

You want to show the individual forces on your free body diagram.

 

[ehabmozart]

Oh! Why do i feel that i am the dumbest person in this world... Dude, I guess i;ve got confused between Restoring force, Force exerted by spring and position of eq. postion... Please clarify cs i understand nothing if u please! Thanks

 

[Doc Al]

Oh! Why do i feel that i am the dumbest person in this world... Dude, I guess i;ve got confused between Restoring force, Force exerted by spring and position of eq. postion... Please clarify cs i understand nothing if u please! Thanks

Only two forces (ignoring air resistance for the moment) act on the mass: The weight, which you know acts down, and the force from the spring, which acts up.

Answer these questions:

Imagine the spring is just hanging there, unstretched. What force does it exert?

Now hang the mass on the spring and bring the mass to the new equilibrium position. What force does the spring exert? What's the net force on the mass?

If the mass moves above the equilibrium point, does the spring force increase or decrease? What happens to the net force on the mass?

If the mass moves below the equilibrium point, does the spring force increase or decrease? What happens to the net force on the mass?

 

[ehabmozart]

Allright, let me take each sitauation. During eqillibrium point.. Net force should be 0, so mg=kx... Now, the sitaution below eqm, Net force is downwards, so y not mg>kx? According to ur question, i really have no idea cs i messed between restoring force and the force of the spring. Btw, this was not a homework task, more than it was a imple question and i needed to know the concept so i would like a straight reply next time .. thanks|!

 

[Doc Al]

During eqillibrium point.. Net force should be 0, so mg=kx...

OK.

Now, the sitaution below eqm, Net force is downwards, so y not mg>kx?

Why do you think the net force is downward?

Only two forces act. The weight is the same. What happens to the spring force as the spring is stretched further downward? Does it increase or decrease? What direction does it act?

 

[ehabmozart]

Whether below or above, the restoring force should be the SAME! beacuase F=-kX... ?

 

[Doc Al]

Whether below or above, the restoring force should be the SAME! beacuase F=-kX... ?

You say the spring force is the same yet it is also proportional to x?

When you stretch the spring downward, which way does the spring pull back on you?

If you stretch it even further downward, how does the spring force change?

 

[ehabmozart]

well, the same whether the same distance x below or above.. Look! I got to anoither doubt which may make this simple.. What is K(x-d) what is d??

 

[Doc Al]

I got to anoither doubt which may make this simple.. What is K(x-d) what is d??

x = amount by which spring is stretched.
d = the position of the equilibrium position below the unstretched position; d = mg/k

The two forces on the mass are:
Upward force of the spring = kx
Downward force of gravity = mg = kd

Net upward force = kx - kd = k(x-d)

Note that when the mass is above the equilibrium point x < d and thus the net force is downward; when below equilibrium, x > d and thus the net force is upward.

 

[ehabmozart]

I really appreciate your effort in making me understand but i am afraid, i am more confused now. To settle this down, am going to give my own words and correct me if i am wrong.. (Please mark which situation you are talking about.. like for example, equillibrium position with /. wothout load)
Anyway, when it is in the new eq. position with the load fx=mg... because the net force has to be zero. I agree with this... Now, mg > k(x-d)... I know that when in motion, the force tries to bring the toy back to the eq. postion.. That's y mg dominates. But what does it actually dominate over.. U said x= amount by which spring is stretched... What do u mean, which point was it stretched from?? Original eq. position or the new one.. In my situation here, when above eq. point, what is x? Then u mentioned d "the position of the equilibrium position below the unstretched position; d = mg/k" Do u meand is the amplitude of the the new system when load is attached... And how did u come to know the formula... d=mg/k.. Thanks a lot by the way for bearing me ( I think am getting more stupid).. :(

 

[Doc Al]

Please stop using 'text speak'--it's annoying.

Anyway, when it is in the new eq. position with the load fx=mg... because the net force has to be zero. I agree with this...

That's because at that point the spring force equals the weight. To find where that point is, just set kx = mg and solve for x. That's the distance we call 'd'.

Now, mg > k(x-d)...

This is confusing. I assume you mean mg > kx, which is true when the mass is above the equilibrium point.

I know that when in motion, the force tries to bring the toy back to the eq. postion.. That's y mg dominates. But what does it actually dominate over..

There are only two forces acting. So it must be the spring force!

U said x= amount by which spring is stretched... What do u mean, which point was it stretched from??

x is the amount the spring is stretched, which is measured from the unstretched position. When the mass is at the equilibrium position, the spring is stretched by an amount 'd'. You can easily solve for 'd' as I indicated above.