Theoretical Physics Question
 

A rope attached to a box and is pulled at an angle of 30* above the horizontal with a force of P.

Is the acceleration of this box the same as in this situation when..

A straight bar is pushed up into the back of a block at 30* above the horizontal with the same force P.


I question this because of don't know if the second situation lowers the Normal Force or if it stays the same. Thanks in advance.


**block, no back

What does it mean when it says pushed into the back of a back?

Well if you look at the vectors of what is applying the force, since they aren't horizontal, they affect the normal (break each force vector into components to determine the y-value). I don't remember how that applys to acceleration (it's been a while!) however I am almost certain the normal force will be different for each, which then in turn will increase or decrease friction.

edit - I think the bar situation will increase the normal, since it is pushing "down" on the box and the normal has to compensate for this "increase in weight".

block, that was a poorly placed typo.


If the forces are broken down into components, then both would, it seems to me, have the same x and y forces (because the same P force is being applied at the same angle). I just question this because I don't know if the bar (or hands, or anything that can push, really) would be any different than the rope. I've been told that it is, but I can't seems to understand why.



think the bar situation will increase the normal, since it is pushing "down" on the box and the normal has to compensate for this "increase in weight". Wouldn't this decrease the normal force because the y component of the force P is also working in the positive y direction?

My advice is to draw it and then draw the componant forces.

The good old free body diagram tells me that the forces are ultimately the same, so that's good enough for me. Thanks!

With the rope, ignoring the x, you are pulling up on the block. With the bar, again ignoring the x, you are pushing down on the block. That just makes the most sense to me, but again it has been a while since I did these problems. Get a box and try it, you'll see what I mean. You push down on the bar and therefore down on the box.

edit - Drew it out and I am almost positive that the bar will add to the over down force on the box while the rope will cancel out some down force and decrease the apparent weight of the box.

I suppose I've been a bit vague in my descriptions, so I've just whipped up a little diagram that should hopefully clear things up.

http://img262.imageshack.us/img262/4...sicsaa9.th.png

Yeah, that's the hell of just using words. They can mean different things to everyone, especially if I know what the diagram looks like already. I'll assume then that the guy who told me that the accelerations caused by these forces were not equal was wrong.

Thanks for the help. [thumbup]

Seems so. Judging by your FBD's, the forces to work out to the same in the end, so it wouldn't matter. I'm sure if you got technical about how high up the box the rope is connected and the bar is pushing it could change, but that seems too complicated for this. I'de hate to have to solve that![rofl]

i'm not sure that they would be the same... if you assume that the box is uniform i.e. the centre of mass is in the middle, then the bar that is pushing upwards is trying to force the opposite edge of the box into the ground - pivoting around the centre of gravity whereas the rope is pulling up and the point of pivot is the far edge therefore trying to raise the whole box off the ground and thus decreasing the frictional force?

or am i just going off on one?!?

http://935.mnetcs.com/thumb/thumb.gif

here is a quick piccy of the bar situation in my head... now that i've drawn it i think i'm talking shite

You have somewhat of a point, but your pivot argument can be applied to the rope too : when the rope pulls up on one end, the other end pivots down into the ground.

However, by pushing up with the pole, the leading edge in contact with the ground might dig in, making it harder to move, whereas pulling up with the rope it's the trailing edge that's in contact with the ground and therefore will not dig in, making the box easier to slide.

However, this requires the modelling of the characteristics of the ground on which the box lies, which IMO is probably out of the scope of the question. However, mentioning this in the answer might yield some extra marks http://discussworldissues.com/forums...lies/wink1.gif

i know that this will also effect the rope part too but as i see it the rope will be pulling upwards on the whole of the box bar the end corner whereas the bar will be pushing up on the side of the box that is on the same side but pushing down on the other half of the box past the centre of mass - is that clear? (i've never been the best at explaining what goes on inside my head...!)

do we know at which point the rope/bar is attached to said box? half way up, top, bottom?

edit: this may help...

http://936.mnetcs.com/thumb/thumb.gif

again, thinking about it its probably a moot point as the average frictional force over the whole area of the bottom of the box will be averaged out...

If you just look at the force exerted into the ground from the edge of the box that is in contact in both scenarios, you will find that the forces are exactly the same, but just on opposite edges. The only difference between the 2 is the edge that's in contact relative to the direction of movement, and this will make a difference if the ground is assumed to be soft.

agreed.

are we any closer to the answer?! [rofl]

Ok, so what if the box is on a conveyor belt? http://discussworldissues.com/forums.../confused1.gif

If the rope and bar are acting on the box in the same horizontal plane, then it should be the same, granted that we're talking about a rigid surface that the box is on.

Wouldn't take off.


[rofl]

It depends on the complexity of the model as to whether the kind of thoughts being put forward here are required.

From the original question, since no specification of point of attachement is given, we should assume a simple model, meaning everyone's free body diagrams are "incorrect". In this simple model, the forces in the diagram should be applied at the centre of mass.

The components of each vector P are then equal whether the box is being pushed or pulled, as they are equivalent to Pcos30 in the positive x-direction and Psin30 in the positive y-direction and should be drawn acting at the centre of the box in the diagram.