# Frictionless block & wedge

1. Aug 24, 2009

### Villhelm

1. The problem statement, all variables and given/known data
(This is Harris-Benson Chap5 Problem 6, the statement reads:
A block of mass m is on a wedge of mass M.
All surfaces are frictionless.
Suppose the wedge is accelerated towards the right by an external force at 5ms^-2 (=a).
What is the acceleration of the block relative to the incline (a')?
The incline is alpha (=37deg) above -ve horizontal:

http://villhelm.webs.com/wedgeblock.png [Broken]

2. Relevant equations
Newtons laws?

3. The attempt at a solution
Ok, I setup the free body diagrams as so:

http://villhelm.webs.com/FBDs.png [Broken]

WEDGE Sum of forces in X (1):
$$Ma - N1sin(\alpha) = MA$$
(where A is the resulting acceleration of the wedge after the block is taken into consideration ...)

BLOCK Sum of forces in X(2):
$$N1sin(\alpha) = m(A + a' cos(\alpha))$$

BLOCK Sum of forces in Y(3):
$$N1cos(\alpha) - mg = -ma' sin(\alpha)$$

I try to remove N1 and isolate A now, for which I get:
>>|======================From (1) + (2) above=====================
>>|$$Ma = MA + m(A + a' cos(\alpha))$$
>>|
>>|$$(m+M)A = ma' cos(\alpha) - Ma$$
>>|
>>|$$A = \frac{ma' cos(\alpha) - Ma}{(m+M)}$$
>>|===========================================================

Now, having an equation in A with no other unknowns, I try to get an equation for a':

>>|===============From (2)*cos(α) - (3)*sin(α) above===================
>>|$$N1sin(\alpha)cos(\alpha) = m(A + a' cos(\alpha)) * cos(\alpha)$$
>>|-
>>|$$N1cos(\alpha)*sin(\alpha) - mgsin(\alpha) = -ma' sin^2(\alpha)$$
>>|=
>>|$$mgsin(\alpha) = mAcos(\alpha) + ma' (cos^2(\alpha) + sin^2(\alpha))$$
>>|=============================================================

>>|===========================================
>>|$$mgsin(\alpha) = m(Acos(\alpha) + a')$$
>>|$$gsin(\alpha) = Acos(\alpha) + a'$$
>>|===========================================

>>|===============substitute in the equation for A==================
>>|$$a' = gsin(\alpha) - Acos(\alpha)$$
>>|
>>|$$a' = gsin(\alpha) - \frac{ma' cos(\alpha) - Ma} {(m+M)}*cos(\alpha)$$
>>|========================================================

>>|===========================================
>>|$$a' * (m+M) = (m+M)gsin(\alpha) -ma'cos^2(\alpha) + Macos(\alpha)$$
>>|
>>|$$a' * (m+M) +ma'cos^2(\alpha) = (m+M)gsin(\alpha) + Macos(\alpha)$$
>>|
>>|$$a'(m+M+mcos^2(\alpha)) = (m+M)gsin(\alpha) + Macos(\alpha)$$
>>|===========================================
Which eventually means I get:

>>|===========================================
>>|$$a' = (m+M)gsin(\alpha) + Macos(\alpha) / (m+M+mcos^2(\alpha))$$
>>|===========================================
Which seems ok in so much as it doesn't have any unknowns on the right that I didn't start with. I don't have the answer to check this against, so I really could use some feedback on whether this is right or wrong.

Sorry for the messy post :uhh: I just wanted to make sure I put everything down in case I was doing some voodoo math. Cheers!

Last edited by a moderator: May 4, 2017
2. Aug 24, 2009

### ideasrule

Are you sure that "a" is supposed to be the acceleration of the wedge if the block didn't exist? By the wording of the problem, I thought the wedge's acceleration with the block was 5 m/s^2.

3. Aug 24, 2009

### Villhelm

Yeah, the small "a" corresponds to the 5 m/s^2.

Basically, the small a is the given 5 m/s^2, associated with the force pushing the wedge, while I used A to represent the unknown acceleration of the wedge after it had also pushed into the block, during the next instant. I should probably have used A to represent it, for clarity. The "a" on the diagram is a typo. I fixed up the first diagram to make it clearer what I am meaning. Cheers for taking the time to look over this for me, it's been hounding me all day :|

By the wording of the problem, I thought the wedge's acceleration with the block was 5 m/s^2

Hmm, I kinda wondered that too, but wasn't sure how to go about dealing with that case, so I figured it was only associated with the initial push. How would I go about dealing with a constant 5m/s^2? Would it be valid to assume the force applied to the block was normal to the incline making the force on the block from the wedge something like:

$$m*\sqrt{a^2+a^2tan^2(\alpha)}$$

Last edited: Aug 24, 2009
4. Aug 25, 2009

### ideasrule

If the wedge's acceleration is given, which I think it is, you don't need to write the equation of motion for the wedge; its motion is already known. Your equations for the block seem right, so if you solve those, you'll get the answer.

An easier way to go is to consider the situation from the wedge's perspective. Gravity exerts mgsin(alpha) on the block parallel to the incline, while the fictitious force mA has the component mAcos(alpha) parallel to the incline. Write out Newton's second law and you're done.

5. Aug 25, 2009

### kuruman

"Suppose the wedge is accelerated towards the right by an external force at 5m/s2" means that 5 m/s2 to the right is the acceleration of the wedge. This problem can be done most easily in the (non-inertial) frame of the wedge. In that frame, you have the usual force of gravity mg down and a fictitious force ma to the left. It is the same analysis as having a stationary wedge and a horizontal force pushing on the block to the left.

** Edit **
That's what ideasrule just said, but I didn't see it.

Last edited: Aug 25, 2009
6. Aug 27, 2009

### Villhelm

Cheers guys! I managed to do the question and get a numeric answer (~1.9m/s^2 down the incline, which seems reasonable). The question really might have been better if it'd appeared in the subsequent chapter, because that one deals with NI frames, which I only realised today when I started reading it ... so I guess I've learnt another thing ... read ahead!