# Homework Help: Finding Normal Reaction between Wedge and Block

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1. Aug 11, 2015

### Titan97

1. The problem statement, all variables and given/known data
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.
2. Relevant equations
None.

3. The attempt at a solution

Let the block a have a velocity vx and vy along x and y-axis with respect to earth. Differentiating w.r.t time, it has an acceleration ay 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?

2. Aug 11, 2015

### bcrowell

Staff Emeritus
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.

3. Aug 11, 2015

### Titan97

Yes. At α=0, N=mg which is correct.

4. Aug 11, 2015

### bcrowell

Staff Emeritus
Units? alpha=90?

5. Aug 11, 2015

### Titan97

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

6. Aug 11, 2015

### bcrowell

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

7. Aug 11, 2015

### Titan97

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

8. Aug 11, 2015

### Qwertywerty

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

9. Aug 11, 2015

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

10. Aug 11, 2015

### bcrowell

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

11. Aug 11, 2015

### insightful

This means that as alpha approaches 90o, ay 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.

Last edited: Aug 11, 2015
12. Aug 11, 2015

### insightful

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

13. Aug 11, 2015

### insightful

Edit, delete.

14. Aug 11, 2015

### Titan97

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.