Find the initial acceleration in a system

[RoloJosh16]

Homework Statement

A cubic box of mass $m$ rests on a horizontal surface and thin uniform rod, of mass $m$ as well is supported against one of the faces of the cube touching it in its centre. The angle between the surface and the rod is $α= 45°$ and there is no friction between any surface. Initially both objects rest as shown in the picture. ¿which is the initial acceleration of the box at the instant the system is released and starts to move?

Relevant Equations

Mechanics

Let´ s call $N_x$ the magnitud of the force between the rod and the box and $N_y$ the magnitud of the force between the rod and the surface.

$N_x = ma_c$
$N_x= ma_r$
$mg-N_y=ma_y$

The following I think is to find a relation between $a_r$ and $a_y$ and that can be found by analysing the initial displacements. Since the centre of mass of the rod has the same acceleration as the box and the length of the box remains the same I can write a relation, I did it but it depends on the length of the rod and do not how to follow.

 

[RoloJosh16]

If I approximate the initial displacements to zero and neglects second order terms then $\Delta y= 2\Delta x$ but I do not know if that is valid

 

[PeroK]

If I approximate the initial displacements to zero and neglects second order terms then $\Delta y= 2\Delta x$ but I do not know if that is valid

You might have to rethink your approach. What else could you use instead of forces?

What are $\Delta x$ and $\Delta y$?

 

[RoloJosh16]

$\Delta x$ would be the initial horizontal displacement of the rod's centre of mass and $\Delta y$ would be the initial vertical displacement.

 

[PeroK]

$\Delta x$ would be the initial horizontal displacement of the rod's centre of mass and $\Delta y$ would be the initial vertical displacement.

I have no idea how you got $\Delta y = 2 \Delta x$. In any case, it can't be right. The motion must depend on the mass of the block.

You need a rethink. What general principles could you apply here?

 

[RoloJosh16]

From mechanics?

 

[PeroK]

From mechanics?

What else? That wasn't a very long think!

 

[RoloJosh16]

What meant was that if I needed a mechanics principle.

 

[PeroK]

What meant was that if I needed a mechanics principle.

Given this is unambiguously a mechanics problem, what else could it be? What have you used in the past - apart from forces - to solve mechanics problems?

 

[RoloJosh16]

torques?

 

[RoloJosh16]

Given this is unambiguously a mechanics problem, what else could it be? What have you used in the past - apart from forces - to solve mechanics problems?

Maybe momentum or energy but I do not how would that help?

 

[PeroK]

Maybe momentum or energy but I do not how would that help?

Yes, of course, momentum and energy and the conservation laws. Thinking about energy and momentum always helps!

I'm going offline now. Have a think about energy and momentum for this problem.

 

[haruspex]

It's valid, but in effect you have found the relationship between the two accelerations and integrated twice. It is more useful to stick with the accelerations.

torques?

yes.

 

[RoloJosh16]

Yes, of course, momentum and energy and the conservation laws. Thinking about energy and momentum always helps!

I'm going offline now. Have a think about energy and momentum for this problem.

Really I do not know how to proceed. I tried writing $W = \Delta K$ but that is not useful, it just adds more variables.

 

[RoloJosh16]

It's valid, but in effect you have found the relationship between the two accelerations and integrated twice. It is more useful to stick with the accelerations.

yes.

If we calculate torques from the points where the two normal intersect (for the rod) and apply angular moment we get $mg= ma_x+ma_y$, is that right?

 

[haruspex]

If we calculate torques from the points where the two normal intersect (for the rod) and apply angular moment we get $mg= ma_x+ma_y$, is that right?

It is not static. There will be angular acceleration.

 

[RoloJosh16]

Well, sorry for being absent a long time. I´ ll go with torques. Writing torques about the center of mass we get:

$\frac {L}{2\sqrt {2}} (N_y - N_x) = I_0 \alpha$

The equation F=ma here was not useful since you added an extra unknown per equation. So I thought that I could relate the forces to the angle the rod made with the surface and therefore to the angular acceleration. I worked out an expression but it is really complicated and I think it is not the way.

 

[haruspex]

I agree with your torque equation.

The equation F=ma here was not useful since you added an extra unknown per equation

But you get such an equation for both bodies, which tells you about their relative acceleration. That is important because you have the fact that they remain in contact, so says something about angular acceleration.