# Mechanics question- block on a wedge

• Mar 14th 2009, 08:40 AM
C.E
Mechanics question- block on a wedge
I was wondering if anyone could offer some help with the following question:

A wedge with mass M rests on a frictionless horizontal table top. A block with mass m is placed on the wedge. There is no friction between the block and the wedge. The system is released from rest.

Calculate the acceleration of the wedge.
Express your answer in terms of M, m,α and g.
The answer I should get is:

http://session.masteringphysics.com/...9%5E%7B2%7D%7D
However I keep getting the following:
-mgsinαcosα
M+mcosαsinα

Does anyone see where I am going wrong? My working is as follows.
I will denote the normal force to the block by n.
n=mgcosα
Therefore the horizontal component of force on the wedge due to the block is given by F=-mgcosαsinα. The combined downward force due to the ramp and block is Mg+mgcosαsinα
Therefore (M+mcosαsinα)a=-mgcosαsinα (By Newtons second law)
Hence a=mgsinαcosα
M+mcosαsinα
I have no idea where I am going wrong so any help would be appreciated.
• Mar 14th 2009, 12:12 PM
Opalg
Quote:

Originally Posted by C.E
I was wondering if anyone could offer some help with the following question:

A wedge with mass M rests on a frictionless horizontal table top. A block with mass m is placed on the wedge. There is no friction between the block and the wedge. The system is released from rest.

Calculate the acceleration of the wedge.
Express your answer in terms of M, m,α and g.
The answer I should get is:

http://session.masteringphysics.com/...9%5E%7B2%7D%7D
However I keep getting the following:
-mgsinαcosα
M+mcosαsinα

Does anyone see where I am going wrong? My working is as follows.
I will denote the normal force to the block by n.
n=mgcosα

I think the error comes right there at the beginning. The normal force is not equal to $\displaystyle mg\cos\alpha$, because the block will have an acceleration in the normal direction as the wedge slides away along the table. In fact, the component of the acceleration in that direction will be $\displaystyle a\sin\alpha$ (where a is the acceleration of the wedge), so Newton's second law gives $\displaystyle n-mg\cos\alpha = ma\sin\alpha$. Therefore $\displaystyle n = m(g\cos\alpha + a\sin\alpha)$. The horizontal equation of motion for the wedge is then $\displaystyle -n\sin\alpha = Ma$, which gives the stated answer for a.
• Mar 14th 2009, 12:48 PM
C.E
Sorry, I do not understand why the normal force would be http://www.mathhelpforum.com/math-he...a0733362-1.gif or why based upon this http://www.mathhelpforum.com/math-he...cdcf54c1-1.gif or even why if http://www.mathhelpforum.com/math-he...615650ef-1.gif, then we have the required result. Could you please elaborate?
• Mar 14th 2009, 02:05 PM
Opalg
Consider the block and the wedge separately. The only forces acting on the block are its weight mg acting downwards, and the normal force n exerted by the wedge. The only forces acting on the wedge are its weight Mg acting downwards, a normal force equal and opposite to n (Newton's third law) exerted by the block, and an upwards normal force (not shown in the picture) exerted by the table.

The wedge slides towards the right with an acceleration a. The motion of the block is a combination of a component $\displaystyle a\sin\alpha$ in the normal direction, together with a relative acceleration down the sloping face of the wedge.

Looking only at the wedge (not the block), the only horizontal force on it is the component $\displaystyle n\sin\alpha$ of n. So the equation of motion in this direction is $\displaystyle n\sin\alpha = Ma$. (I seem to have reversed the sign of a somewhere along the line, so the acceleration is going to lose its negative sign.)

Now looking only at the block (not the wedge), the equation of motion in the normal direction says that the forces acting in that direction, namely $\displaystyle mg\cos\alpha-n$, must produce the acceleration $\displaystyle a\sin\alpha$. (If the block does not have that component of acceleration in that direction then it will not stay in contact with the wedge.) So Newton's second law says that $\displaystyle mg\cos\alpha-n = ma\sin\alpha$.

I don't know if that makes it any clearer. Apart from drawing the picture (and changing the sign of a), I seem to have more or less repeated my previous comment. I don't see what more I can give by way of explanation.
• Mar 14th 2009, 02:50 PM
C.E
Thanks that did help quite a lot, I know understand everything you said apart from why the force acting on the block in the normal direction is given by http://www.mathhelpforum.com/math-he...d2ee67b9-1.gif. I do not see where the -n comes from.
• Mar 14th 2009, 02:56 PM
mr fantastic
Quote:

Originally Posted by C.E
Thanks that did help quite a lot, I know understand everything you said apart from why the force acting on the block in the normal direction is given by http://www.mathhelpforum.com/math-he...d2ee67b9-1.gif. I do not see where the -n comes from.

The weight force has a normal component. This component is in the opposite direction to the normal reaction force ....
• Mar 19th 2009, 09:46 AM
C.E
Question parts 2 and 3.
I was wondering if anyone could give me some hints as to how I am supposed to do part 3 of this question.
2.Calculate the horizontal component of the acceleration of the block.
Express your answer in terms of http://session.masteringphysics.com/render?tex=M, http://session.masteringphysics.com/render?tex=m, http://session.masteringphysics.com/render?tex=%5Calpha, and constant http://session.masteringphysics.com/render?tex=g.

3.Calculate the vertical component of the acceleration of the block.Express your answer in terms of http://session.masteringphysics.com/render?tex=M, http://session.masteringphysics.com/render?tex=m, http://session.masteringphysics.com/render?tex=%5Calpha, and constant http://session.masteringphysics.com/render?tex=g.

Note I have done part 2 and correctly got the answer ( I have added it to my post as I think it may be needed for part 3). The answer to part 2 is:
http://session.masteringphysics.com/...9%5E%7B2%7D%7D
Anyway for part 3 I keep getting the wrong answer, could someone please explain how to do it?
By the way the answer you should get is:
http://session.masteringphysics.com/...D%5E%7B2%7D%7D
• Mar 20th 2009, 06:54 AM
Opalg
Quote:

Originally Posted by C.E
I was wondering if anyone could give me some hints as to how I am supposed to do part 3 of this question.
2.Calculate the horizontal component of the acceleration of the block.
Express your answer in terms of http://session.masteringphysics.com/render?tex=M, http://session.masteringphysics.com/render?tex=m, http://session.masteringphysics.com/render?tex=%5Calpha, and constant http://session.masteringphysics.com/render?tex=g.

3.Calculate the vertical component of the acceleration of the block.Express your answer in terms of http://session.masteringphysics.com/render?tex=M, http://session.masteringphysics.com/render?tex=m, http://session.masteringphysics.com/render?tex=%5Calpha, and constant http://session.masteringphysics.com/render?tex=g.

Note I have done part 2 and correctly got the answer ( I have added it to my post as I think it may be needed for part 3). The answer to part 2 is:
http://session.masteringphysics.com/...9%5E%7B2%7D%7D
Anyway for part 3 I keep getting the wrong answer, could someone please explain how to do it?
By the way the answer you should get is:
http://session.masteringphysics.com/...D%5E%7B2%7D%7D

You just have to write down the equation of motion for the block, in the vertical direction. Referring to the picture again, the vertical forces on the block are the upward component $\displaystyle n\cos\alpha$ of the normal force, and the downwards weight mg of the block. So the net upwards force is $\displaystyle n\cos\alpha - mg$, and the equation of motion says that this is equal to mc, where c is the vertical acceleration of the block.

But we know that $\displaystyle n = \frac{Ma}{\sin\alpha}$ and $\displaystyle a = \frac{mg\sin\alpha\cos\alpha}{M+m\sin^2\alpha}$. So $\displaystyle n = \frac{Mmg\cos\alpha}{M+m\sin^2\alpha}$, and the equation of motion is

$\displaystyle mc = \frac{Mmg\cos^2\alpha}{M+m\sin^2\alpha} -mg = \frac{Mmg\cos^2\alpha -mg(M+m\sin^2\alpha)}{M+m\sin^2\alpha}$. Simplify that (using $\displaystyle 1-\cos^2=\sin^2$) and you'll get the answer.