Finding Normal Reaction between Wedge and Block

[Titan97]

Homework Statement

A block of mass m is kept on a smooth wedge of angle α and mass M which is in turn kept on a smooth floor.
When the system is released from rest, calculate the force exerted by block on the wedge.

Homework Equations

None.

The Attempt at a Solution

Let the block a have a velocity v_x and v_y along x and y-axis with respect to earth. Differentiating w.r.t time, it has an acceleration a_y along y-axis. For the block to be in contact with the wedge, $a_y=asin\alpha$
Let $N$ be the normal reaction Force between wedge and block..
From F.B.D of wedge, $Nsin\alpha=Ma$ , so $N=Macosec\alpha$
From F.B.D of block, $mgcos\alpha-N=ma_y$ and $mgsin\alpha=ma_x$
hence, $mgcos\alpha-N=masin\alpha$ and $a=\frac{mgcos\alpha}{msin\alpha+Mcosec\alpha}$
Therefore, $$N=\frac{mgcos\alpha}{\frac{m}{M}sin^2\alpha+1}$$
Is this correct?

 

[bcrowell]

There are some standard things you should automatically do to check yourself on a problem like this: (1) check the units, and (2) check the dependence on the variables. Do you know how to do these two things, and have you done them? In more detail, I would suggest that in #2 you check the two special cases $\alpha=0$ and $\alpha=90$ degrees.

By the way, "normal reaction force" is presumably terminology that you picked up from your text or your teacher, but IMO it's bad terminology. There is no such thing as a reaction force. Newton's 3rd law is symmetric, and neither force plays the role of action or reaction. Also, there's a common misconception that normal forces are somehow specially linked to the 3rd law or are always reaction forces. Newton's third law applies to all types of forces in mechanics, not just normal forces.

 

[Titan97]

Yes. At α=0, N=mg which is correct.

 

[bcrowell]

Units? alpha=90?

 

[Titan97]

At α=90°, N=0. Yes. The dimension of the final answer is correct.

 

[bcrowell]

Looks right to me. Another interesting special case to check is the one in which $M\rightarrow\infty$.

 

[Titan97]

At M→∞, N→mgcosα?
If the wedge is infinitely heavy, It wouldn't move (right?).
But why is N=mgcosα?

 

[Qwertywerty]

Post edited : Found my mistake . Your calculations look correct .

 

[Titan97]

Actual question is to check if N is greater than or less than mgcosα. The answer given and the answer i got is same.

N=$\frac{mgcos\alpha}{\frac{m}{M}sin^2\alpha+1}$

The above value is clearly less than $mgcos\alpha$
I think there is no problem in my answer. What did you get?

 

[bcrowell]

At M→∞, N→mgcosα?
If the wedge is infinitely heavy, It wouldn't move (right?).
But why is N=mgcosα?

I think that's what you get if you solve the problem with the wedge held immobile (so that $a_y=0$).

 

[insightful]

For the block to be in contact with the wedge, ay=asinαa_y=asin\alpha

This means that as alpha approaches 90^o, a_y approaches a.
But, shouldn't it approach g?

Edit: I see the confusion. Usually "with respect to the earth" means y is vertical and x is horizontal.

 

[insightful]

Let the block a have a velocity vx and vy along x and y-axis with respect to earth.

Oh, do you mean "with respect to the top of the wedge" (as drawn)?

 

[insightful]

Edit, delete.

 

[Titan97]

Oh, do you mean "with respect to the top of the wedge" (as drawn)?

I fixed the x-axis as the surface of the wedge. At that instant, vx,vy are the block's velocity components w.r.t earth.