# Homework Help: Mass on accelerating wedge

1. Dec 31, 2007

### Fermat

1. The problem statement, all variables and given/known data
A 45 degree wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge. Find its acceleration (gravity is downwards),

2. Relevant equations

3. The attempt at a solution
I answered the above question on another forum, but got the wrong answer, apparently. Here's my working. Where did I go worng?

Suppose that the wedge is stationary. Then the accelerating force, down the slope, is mgsin45. So the acceleration of the small mass, down the slope, is gsin45.

If the wedge is now moving with a (constant) accln of A, then that accln can be added vectorially to the "static" accln of the small mass.

http://img223.imageshack.us/img223/1896/wedge1ub9.th.jpg [Broken]

a is the resultant accln of the two component acclns, and (vectorially), a = A + gsin45.

Your question isn't clear in which direction the force is acting wrt to the orientation of the wedge. The force could be in the same direction as the wedge is pointing, or in the opposite direction, in which case the vector diagram is like this,

http://img223.imageshack.us/img223/299/wedge2pc6.th.jpg [Broken]

If you wish to evaluate the accln, then use the cosine rule.

a² = A² + (gsin45)² - 2A.gsin45.cos(45 or 135)

Now, the OP told me that if A = 3g, then the answer is g. But I can't see how that answer is got. Can someone explain

Last edited by a moderator: May 3, 2017
2. Jan 1, 2008

### metalInferno

see when the mass slides down , the diagonal component of it weight is $$mg sin(\theta)$$ whereas the pseudo force component acting diagonally up is $$mAcos(\theta)$$.
even then if u put a=3g , then the answer is $$-g\sqrt{2}$$.

3. Jan 1, 2008

### metalInferno

one more thing if by any chance the acc is greater than g , the block tends to move upwards rather than downwards , because it actually stops when a=g . plz feel free to point out errors if i made any in saying this . so as far as i know the answer can be anything but positive . but if u r only talking about the magnitude then ... can u at least post a pic of the prob , in case i made a mistake in understanding the question ....

4. Jan 1, 2008

### Fermat

I don't know what the actual digram may have looked like - I only got the question text - but I imagine that the two options may have looked like the below.

Wedge moving to the left
http://img215.imageshack.us/img215/2241/wedgetotheleftxw3.th.jpg [Broken]

Wedge moving to the right
http://img215.imageshack.us/img215/3333/wedgetotherightbo2.th.jpg [Broken]

If A = 3g, then the acceln, along the slope is,as you say, of magnitude $g\sqrt{2}$. But the small mass still has an acceln component perpindicular to the slope's surface, which gives it a resultant accln, relative to the ground along which the wedge is moving.

I assumed that this was the acceleration being looked for - one that was relative to the ground. But whether relative to the slope's surface or to the ground, I still can't see how an accln of g (for A = 3g) comes about ???

Last edited by a moderator: May 3, 2017
5. Jan 1, 2008

### arildno

Go into an accelerated frame, accelerating with A in the horizontal direction with respect to an inertial frame with non-rotated coordinate axes with respect to the accelerated frame..

Let us set up the relations between unit vectors, with i being unit vector in horizontal direction, j in vertical direction, n in normal direction, and t in tangential dirction:
$$\vec{t}=\cos\theta\vec{i}-\sin\theta\vec{j},\vec{n}=\sin\theta\vec{i}+\cos\theta\vec{j}$$
Or, alternatively:
$$\vec{i}=\cos\theta\vec{t}+\sin\theta\vec{n},\vec{j}=-\sin\theta\vec{t}+\cos\theta\vec{n}$$
Thus, in the accelerated frame, we have:
$$-mA\vec{i}-mg\vec{j}+\vec{N}=m\vec{a}_{rel}$$
Now, $$\vec{a}_{rel}$$ is strictly tangential, and our component force equation is then:
$$-mA\cos\theta+mg\sin\theta=ma_{rel}$$,
$$a_{rel}=g\sin\theta-A\cos\theta$$

Thus, its absolute acceleration is:
$$\vec{a}=A\vec{i}+a_{rel}\vec{t}=A\sin\theta\vec{n}+g\sin\theta\vec{t}$$
whereby the acceleration a is given as:
$$a=\sqrt{A^{2}+g^{2}}\sin\theta$$
Note that in the special cases theta=0 and theta=90 deg, in the first, the frictionless block just slides under the now stationary mass, whereas in the 90 deg case, the mass experiences free fall downwards and A acceleration horizontally.

At A=3g, we have that $$a=\sqrt{5}g$$, so neither OP or you other two are correct.

When A=g, with theta=45, then its acceleration is evidently A, that is g, which we see from my formula.

Last edited: Jan 1, 2008
6. Jan 1, 2008

### Fermat

So, my OP had a wrong answer in his book, looks like. Thanks for confirming that.