Physics questions thread

[#1procrastinator]

If we stack two blocks whose surfaces are completely smooth, one on top of another and push the bottom one (floor is also smooth), would the force required to give the bottom block a certain acceleration be the same as if there were nothing on it?

e.g. If we have 2kg block on the bottom and a 10kg block on top of it and pushed the 2kg block with a force of 10N, would the acceleration be 5m/s^2?

[Homer]

I guess even if the surfaces are completely smooth the force required would be greater, due to the gravity force acting downwards, however this is just a wild guess :p

[Quantum.Mechanic]

It depends on whether you want to move the whole object as one or not. But if they are both smooth, you can't shift the top block if you provide a force only to the bottom block. Otherwise only the bottom block would move, the top block would fall off..

This is shown in mythbusters, if one can move something fast enough, you are able to only move the object on the bottom. I remember the episode Tabletop Chaos, where they try to move only the table cloth without shifting the tableware on top of the table. If one can provide sufficient force/speed/acceleration, then one can shift the tablecloth without shifting the tupperware and the like on the table.

Its not exactly the same thing, but the concept is pretty much the same. The acceleration would have to be 5m/s^2, assuming the force is provided on the bottom block ONLY

[#1procrastinator]

I guess even if the surfaces are completely smooth the force required would be greater, due to the gravity force acting downwards, however this is just a wild guess :p

That's what I thought made intuitive sense too but the added weight would just increase the total normal force which would increase the frictional foce making it harder to push. But in this imagined case, there's zilch friction!

It depends on whether you want to move the whole object as one or not. But if they are both smooth, you can't shift the top block if you provide a force only to the bottom block. Otherwise only the bottom block would move, the top block would fall off..

This is shown in mythbusters, if one can move something fast enough, you are able to only move the object on the bottom. I remember the episode Tabletop Chaos, where they try to move only the table cloth without shifting the tableware on top of the table. If one can provide sufficient force/speed/acceleration, then one can shift the tablecloth without shifting the tupperware and the like on the table.

Its not exactly the same thing, but the concept is pretty much the same. The acceleration would have to be 5m/s^2, assuming the force is provided on the bottom block ONLY

So in an ideal situation where all surfaces are completely frictionless, it wouldn't matter how much mass you stack on right, the force required to accelerate the bottom mass (for some a) would always be the same?

The reason I'm asking is because I'm working on this problem:

A 'pedagogical machine' is illustrated in the link below. All surfaces are frictionless. What force F must be applied to M1 to keep M3 from rising or falling.

http://hep.physics.wayne.edu/~harr/courses/5200/w99/problem6-4.gif

And I want to know if the acceleration of the system would include M2 (might be heading in the wrong direction here...)

[Quantum.Mechanic]

See the problem is in an ideological situation that exactly would occur, due to lack of friction. You will see the top block move as it will fall off, but it will not accelerate, it will just fall due to gravity once the lower block has moved away...

You would have to include the mass of m2 in the equation. This would be because of gravity, and the tension between the two objects. If you have the tension between the 2 objects, you have to consider it as a whole mass, because moving m2, will also move m3.

Do you have the exact problem, or is this solving equations with just unnumbered variables.

[#1procrastinator]

thanks.

when M1 accelerates, M3 would accelerate horizontally at the same rate right? so the force from M1 on M3 would be different than the external force on the system?  and since M3 is attached to M2, then it pulls in M2 making it part of the system...

yeah, that's the actual question. very few numbers in this book which I much prefer :]

[Quantum.Mechanic]

I assume M3 would accelerate at the same rate forward, since its stuck in the crevice thingamaggige.
If they are touching each other, the friction shouldn't affect the force provided, it should rather be the same. The only thing that is working on m3 is gravity. And since M2 and m3 are connected, it should be rather part of the whole system.
So by moving M1, M2 and M3 should move due to gravity downwards, rather than sliding off, due to the tension in the rope.
I think thats my interpretation of it. I should read more about Physics, rather than the simple IB and First Year Uni ones.

[#1procrastinator]

Ok, i think i got the it now. The acceleration of M2 to the right is T/M2 so the acceleration to the left should be equal to this so that M3 doesnt fall. So i think it's ok to consider relative acceleration for this and say as M1 accelerates to the right, M3 would accelerate to the left (from M1s point of view though it is M1 that's feeling the force). So the acceleration of M1 is F/(M1+M2+M3) so then equate that and the above equation and solve for F which gives

Have no idea if that is correct though cause there is no solutions in the book lol. Only an hint which says 'for equal masses, F=3Mg' which i just realised answered one my questions above :p

The book is called 'an introduction to mechanics' by kleppner and kolenkow if you're interested. It's meant for a first year course but i think it's hard to learn from without other resources (i'm mainly learning from one of those standard bigass first-year textbooks. Quite a few concepts are .the problems are great though, really makes you think about the physixs carefully rather than rely on mathematical trickery.

Anyway, i have no idea yet how to do this next question which applies to the same pedagogical machine in the previous one.

Consider the pedagogical machine of the last problem in the case where F=0. FInd the acceleration of M1

[PlainElegy]

If this thread is still alive...in the context of explaining the particle model, why is it that the motion of an object is not influenced by the "details" of the object's size or shape? I thought friction, air resistance, etc would be considered? Or is it the fact that we are ANALYSING motion that is already happening as opposed to finding an aspect of the motion (eg final velocity)?

[saradom136]

Can anyone help with this?  

A horizontal 0.65 m rod, weighing 3.6 kg, has a pin protruding from the centre of one end and a string with a 0.66 kg box hanging from the other end. Calculate how much torque the pin must exert to keep the rod from rotating.
[Hint: Assume the string is massless. g = 9.80 m/s2]

[Conic]

Can anyone help with this?  

A horizontal 0.65 m rod, weighing 3.6 kg, has a pin protruding from the centre of one end and a string with a 0.66 kg box hanging from the other end. Calculate how much torque the pin must exert to keep the rod from rotating.
[Hint: Assume the string is massless. g = 9.80 m/s2]

First find the downward torque due to gravity (Using pin as the pivot): 



This is equal to the upward torque, so the pin provides a torque of 16Nm.

I found this question annoying, because they never told you what point to use as the pivot, but this method seemed to get the right answer.

[saradom136]

Thanks 

[hobbitle]

Wasn't sure where to put this question but I just have a little Q about units.

Say I want to know what units something will end up in, I know you can 'calculate' using just units (but without actual values assigned to the variables).  Say I have a formula and one of the variables in it is squared - do I square the units of it, too?

[nerdgasm]

Wasn't sure where to put this question but I just have a little Q about units.

Say I want to know what units something will end up in, I know you can 'calculate' using just units (but without actual values assigned to the variables).  Say I have a formula and one of the variables in it is squared - do I square the units of it, too?

Just as a bit extra, what happens when we take derivatives? For standard derivatives, say  , the units for these are typically . For example, we have , and the units of acceleration as stated earlier, were . Why is this? Well, let's write our standard derivatives in 'English-speak'.

If I wanted to describe , I would say "the rate of change of x, with respect to t". Or in other words, this describes "the ratio of the rates of change of x and t". In other words, it's like having (which is really just a non-infinitesimal version of , so this kinda makes sense), which would have units of

Now, when we take a higher derivative like , this is really like , so it's like "the ratio of the rates of change of (dx/dt) and t". In other words, we now have something like  , which can be treated like a fraction to give , which explains the units for acceleration.

So yeah, hopefully this helps a bit.

[hobbitle]

Thanks BA, and to nerdgasm for the expanded answer! That was really helpful!