Recent content by RoloJosh16

It is not affected(?)

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...

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?

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

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

torques?

What meant was that if I needed a mechanics principle.

From mechanics?

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

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

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...

Thanks, you have helped me a lot :)

Though in this case it is like a combined acceleration. The acceleration of the masses depend on the acceleration of the wedge and both would accelerate immediately to the needed value.

Ahhh thank you I already found the answer with that. When thinking about this problem I thought that there had to happen some time since the ball is placed in top of the wedge until the wedge acquires the needed acceleration but would it not be strange that acquires it immediately?

I´ d like to relate those forces to the acceleration of the wedge, but I do not know its mass. My statement of the minimum acceleration is correct?