Easy physics question!

[Sien]

Alright awesome forum people! I need your help!

This has been a long week, and I might not be thinking as well as I would like, it might explain any herp derp mistakes I made.

Me and my 11th grade physics teacher are in a bit of a stand-still. We are having a little disagreement on a problem we did today.

We have a bloc (2kg) on an inclined plane (25 degrees, angled up). Supporting this block (bellow, so the block does not slide off the plane.) is a spring with a strength of 50n/m. The friction coefficient (cinematic, we neglect the other) is 0.2. This system as a whole is at rest.

Now, we add a constant force pulling at the block with 10N, angled 10 degrees up relative to the inclined plane.

The question is: What is the speed of the block when it has moved 20 centimetres?

But my problem is that the forces seem to balance out when the spring is stretched about 19.3 centimetres... there does not seem to be a way for the block to actually get to 20 centimetres in the fist place.

So, what did I do wrong?

[Dopefish]

Science forum rather then school for this sort of thing typically. (edit: And now it's been moved.)

I haven't done any calculations, but remember that a force balance doesn't mean that the object isn't moving, just that it isn't accelerating. Hence assuming your calculation that the forces are balanced after it's stretched 19.3 cm is correct, then it seems likely that at that point it will still have enough velocity to continue travelling to the 20 cm mark. The net force will be pointing opposite it's motion at this point of course, slowing it down and eventually pulling it back, but until that deceleration has occured it'll keep on moving.

[yurell]

As Dopefish said, forces only take into account your acceleration.

F = ma = 0
=> a = 0
=> dv/dt = 0
=> v = const.

So at the point where the forces balance you have constant velocity, not necessarily zero. When the force starts pushing it back up the slope, it will begin to slow down, reaching its maximum displacement when v = 0.

[Sien]

Ahhhh... there we go, that makes much more sense then the "variable forces" my teacher was rambling on about.

Thanks a lot guys!

[LaserGuy]

Alright awesome forum people! I need your help!

This has been a long week, and I might not be thinking as well as I would like, it might explain any herp derp mistakes I made.

Me and my 11th grade physics teacher are in a bit of a stand-still. We are having a little disagreement on a problem we did today.

We have a bloc (2kg) on an inclined plane (25 degrees, angled up). Supporting this block (bellow, so the block does not slide off the plane.) is a spring with a strength of 50n/m. The friction coefficient (cinematic, we neglect the other) is 0.2. This system as a whole is at rest.

Now, we add a constant force pulling at the block with 10N, angled 10 degrees up relative to the inclined plane.

The question is: What is the speed of the block when it has moved 20 centimetres?

But my problem is that the forces seem to balance out when the spring is stretched about 19.3 centimetres... there does not seem to be a way for the block to actually get to 20 centimetres in the fist place.

So, what did I do wrong?

It's possible that your 10N force is just not strong enough to move overcome the friction and move the block. There is no reason that it has to move. If the system is initially at rest, and the forces are always zero, then the block won't move. At a quick glance, I'd suggest it probably won't.

Gravity down the incline is mgsin(25). This works out to ~9N. Your force is only 10N, and isn't pointing in the same direction. Plus you have friction to overcome, and the restoring force due the spring. I do not think this system will move at all.

[mfb]

In the initial position, the spring applys a force of 0N, if it is not already compressed (It is not in the text, so I don't know). At 20cm, the force increased by 10N. This might fit with the description of zero net force there.

[LaserGuy]

In the initial position, the spring applys a force of 0N, if it is not already compressed (It is not in the text, so I don't know). At 20cm, the force increased by 10N. This might fit with the description of zero net force there.

If the forces are zero at the initial position, and it is initially at rest, it will never reach 20 cm.

[mfb]

>> If the forces are zero at the initial position
Just the force of the spring.
Did you actually calculate that system? Or how do you get to your conclusion?

[Dopefish]

The correct course of action at this point is clearly to use the lagrangian.

Extreme overkill and apparent witchcraft is totally the best way to do first year high school physics.

(I'm a bit preoccupied at the moment, but, someone should. )

[yurell]

How do you use the Lagrangian when you're subject to friction forces?

[eta oin shrdlu]

The way I interpret the problem is that the block's initial position is set so that the spring's restoring force (upward along the inclined plane) is balancing the component of the gravitational force downward along the plane. This is why mention is made that the static coefficient of friction is being neglected. This is a little bit contrived, since in reality the static friction would allow a range of equilibrium positions, but that's the way I read the initial setup.

So when the 10N force is applied there are now five forces in the free-body diagram: the applied 10N force, the spring's restoring force, the gravitational force, the normal force, and the kinematic-frictional force. The first two initially have components upward along the plane, while gravity and friction initially act downward.

With this assumption I find that the block will begin to move, and will in fact manage to move 20cm from its initial position. Finding the velocity at this point can be done either with energy conservation or kinematic simple-harmonic-motion equations; I got the same answer both ways (... eventually...), so that was nice.

[legend]

How do you use the Lagrangian when you're subject to friction forces?

Even more of a reason to do it
It can actually be done in some specific cases. E.g. consider L=\frac{1}{2}\dot{q}^2e^{\lambda q}

[Dopefish]

How do you use the Lagrangian when you're subject to friction forces?

Even more of a reason to do it
It can actually be done in some specific cases. E.g. consider L=\frac{1}{2}\dot{q}^2e^{\lambda q}

This is the correct attitude to take on the matter.

(I may have technically overlooked the friction component, and the lagrangian may not actually be super useful in solving this particular problem even if there was no friction. However, the Lagrangian is still my suggested course of action towards all physics problems intended for an audience who almost certainly has never heard of it. As such, it is very rarely a good suggestion, but I think it's always an awesome suggestion. )

[yurell]

That's why I asked 'how' I've only ever seen the Lagrangian used in frictionless environments, and otherwise have no idea how to encode forces into it.

[starslayer]

Finding the velocity at this point can be done either with energy conservation or kinematic simple-harmonic-motion equations; I got the same answer both ways (... eventually...), so that was nice.

You could also just integrate the forces to get the velocity (solve for a, integrate wrt x). Chain rule ho!

I do like Dopefish's suggestion of using the Lagrangian, though. And yes, it is possible to include ordinary kinetic friction in it. Hell, the second link goes through a virtually identical problem. You would also need to use that approach to include the constant force pulling the block up the slope.

[LaserGuy]

>> If the forces are zero at the initial position
Just the force of the spring.
Did you actually calculate that system? Or how do you get to your conclusion?

From the description, I would assume that the spring force is zero in the initial position, since no extension is given. So the only forces that you need to consider are the gravitational force driving it down the incline, the friction force resisting motion in whatever direction, and the applied force. It actually looks like it should go down the incline.

Gravitational force is mgsin25 = 8.28N
Applied force along incline is Fsin35 = 5.73N
Normal force is mgcos25 - Fcos35 = 9.57N
Friction force is pointing up the incline with a force of uN = 1.91N
Net force is therefore 8.28 - 5.73 - 1.91 = 0.64N down the incline.

[starslayer]

Uh, the force along the incline is Fcos(10) (and the applied force's contribution to the normal is thus Fsin(10)), which is just about 9 N. Also, since the kinetic friction only applies if the block is moving, the system is initially at rest, and the applied force on its own is enough to counteract gravity, the block just won't move, as you initially concluded. Friction cannot cause something to move backwards.

[Dopefish]

My interpretation of the problem has the initial spring force [along the incline] exactly balancing the gravitational force (since the system is at rest initially) [along the incline], so 10N*cos(10) would be the net force up the incline when that additional pulling force is first applied by my reading. The spring (intially being compressed from it's equilibirum position) would decompress and then begin pulling as the spring stretched past it's equiblibrium position. I don't believe knowing where that equilibrium position is actually matters, since you're given the spring constant.

(This is the same interpretation as eta described above.)

edit: Now strictly speaking theres no reason in the problem why it couldn't just be starting from some unstable starting point where the spring is already horribly stretched/compressed and the problem description starts when the object is just at a point during some oscillations where it happens to be at rest, but that'd be a terrible problem. At the high school level homework problems are typically solvable. (Unfortunately, that's not always the case. My advanced classical mechanics prof had a terrible habit of making up problems that just completely didn't work, even with the full might of the Lagrangian [and bigger guns still] being thrown at it. )