Two moving objects - box and cart

AI Thread Summary
The discussion revolves around a physics problem involving a moving cart and a box on it, where the cart's velocity is defined by a function of time. Key points include the need to determine the dimensions of constants involved, the relationship between acceleration and friction coefficients, and the conditions under which the box begins to move relative to the cart. Participants express confusion about the timing of the box's movement and the implications of the cart's acceleration on the box's behavior. The conversation highlights the complexities of static and kinetic friction in relation to the motion of the box as the cart accelerates.
titansarus
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Homework Statement


We have a cart (It is very long) and a box on it.They are at first stationary but at ##t=0##, cart begins to move with ##v## such that:
##v = \alpha t^\beta ; t\leq t_0## and ##\dot v = c; t>t_0 ##
where ##\alpha ,\beta ,c## are constants.

At the time ##t1 (t_1<t_0)## the box is at the verge of movement (I don't know the exact translation of this in English, it means that at ##t_1 + \delta t## (very small ##\delta t>0##) it will move). At the (t_2 > t_0) it will get stationary relative to the cart (both move at the same speed).

I)Find dimension of ##\alpha,\beta,c##
II)Find ##c## in term of ##\alpha, \beta , t_0##
III)Find ##\mu_s ,\mu_k## (Static and Kinetic coefficient of friction) between box and cart.
IV) Find the distance the box traveled on the cart between (t_1 to t_2).

The Attempt at a Solution


I) For the first part, we can simply find that ##\beta## is dimensionless, ##v = L / T##so ##alpha = L * T^ -(\beta +1) ## also c is ##L / T^2##

II) we can say that at ##t = t_0## the accelerations must get equal (Because it's physics, not some strange math functions) so ##\alpha \beta t_0^{\beta-1} = c##

III) We can say that at ##t_2## we have static friction and acceleration of cart and box are equal, so ##\mu_s g = c## and we get ##\mu_s = c/g##. I don't know how to find ##\mu_k## and also how to solve part IV .
 
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titansarus said:
the accelerations must get equal (Because it's physics, not some strange math functions)
You do seem to need to assume this to answer the question, but I don't buy it. There is no physics rule that says accelerations must vary continuously. When a sliding box comes to rest on a frictional floor, its acceleration goes through a step function. Ok, it might become continuous if you take into account that the box is not rigid, but you have no basis for doing that here.
titansarus said:
We can say that at t2 we have static friction and acceleration of cart and box are equal,
Same problem. We can only say they are equal after the speeds have become equal. Prior to that, the box will have been accelerating faster than the cart by some constant difference.

One thing I don't get... how can t1 be > 0?
Edit: Does it mean that at t1 it starts to move relative to the cart?
But if so, we must have β>1 and αβt0β-1>c.
 
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haruspex said:
You do seem to need to assume this to answer the question, but I don't buy it. There is no physics rule that says accelerations must vary continuously. When a sliding box comes to rest on a frictional floor, its acceleration goes through a step function. Ok, it might become continuous if you take into account that the box is not rigid, but you have no basis for doing that here.

Same problem. We can only say they are equal after the speeds have become equal. Prior to that, the box will have been accelerating faster than the cart by some constant difference.

One thing I don't get... how can t1 be > 0?
Edit: Does it mean that at t1 it starts to move relative to the cart?
But if so, we must have β>1 and αβt0β-1>c.

First about ##t_1##; the whole movement and question started at t=0. So I think it is obvious that this must be greater than 0.

About the second part, I think it starts to move relative to inertial frame at that time and not the Cart. the issue is that it's all the question gives and nothing more. I don't know the exact translation of the verge of movement in English, but for example think you are pushing something on floor with static friction coefficient ##\mu_s##, when ##F = mg\mu_s## I say it is at the verge of movement and when ##F > mg\mu_s## it moves. Except from this, It is all of the givings and assumptions of the question. It doesn't have anything else. It is from "General Physics 1" Midterm exam of the best technology university in Iran (Sharif University of Technology).

I also added the picture of the question, although it doesn't have anything useful.

physics.png
 

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titansarus said:
First about t1; the whole movement and question started at t=0. So I think it is obvious that this must be greater than 0.
I agree it implies t1>0, but I do not see how the box can remain stationary in the lab frame after the cart has started to move.
If instead we suppose it means on the verge of slipping, we can find μs in terms of t1, α and β.

We are not told whether β is more or less than 1. If less, the acceleration of the cart is infinite at t=0 and the box must start to slip immediately. So let us suppose β>1.
For the box to slip at all, the peak acceleration of the cart must exceed what the static friction can deliver. If the acceleration is continuous through t0, as you presumed, the acceleration never falls below that peak, so the speeds will never match.

Have you tried to sketch the acceleration-time graph of the cart?
 
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