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Villhelm
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Homework Statement
(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
Homework Equations
Newtons laws?3. The Attempt at a Solution
Ok, I setup the free body diagrams as so:
http://villhelm.webs.com/FBDs.png
WEDGE Sum of forces in X (1):
[tex]Ma - N1sin(\alpha) = MA[/tex]
(where A is the resulting acceleration of the wedge after the block is taken into consideration ...)
BLOCK Sum of forces in X(2):
[tex]N1sin(\alpha) = m(A + a' cos(\alpha))[/tex]
BLOCK Sum of forces in Y(3):
[tex]N1cos(\alpha) - mg = -ma' sin(\alpha)[/tex]
I try to remove N1 and isolate A now, for which I get:
>>|======================From (1) + (2) above=====================
>>|[tex]Ma = MA + m(A + a' cos(\alpha))[/tex]
>>|
>>|[tex](m+M)A = ma' cos(\alpha) - Ma[/tex]
>>|
>>|[tex]A = \frac{ma' cos(\alpha) - Ma}{(m+M)}[/tex]
>>|===========================================================
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===================
>>|[tex]N1sin(\alpha)cos(\alpha) = m(A + a' cos(\alpha)) * cos(\alpha)[/tex]
>>|-
>>|[tex]N1cos(\alpha)*sin(\alpha) - mgsin(\alpha) = -ma' sin^2(\alpha)[/tex]
>>|=
>>|[tex]mgsin(\alpha) = mAcos(\alpha) + ma' (cos^2(\alpha) + sin^2(\alpha))[/tex]
>>|=============================================================
>>|===========================================
>>|[tex]mgsin(\alpha) = m(Acos(\alpha) + a')[/tex]
>>|[tex]gsin(\alpha) = Acos(\alpha) + a'[/tex]
>>|===========================================
>>|===============substitute in the equation for A==================
>>|[tex]a' = gsin(\alpha) - Acos(\alpha)[/tex]
>>|
>>|[tex]a' = gsin(\alpha) - \frac{ma' cos(\alpha) - Ma} {(m+M)}*cos(\alpha)[/tex]
>>|========================================================
>>|===========================================
>>|[tex]a' * (m+M) = (m+M)gsin(\alpha) -ma'cos^2(\alpha) + Macos(\alpha)[/tex]
>>|
>>|[tex]a' * (m+M) +ma'cos^2(\alpha) = (m+M)gsin(\alpha) + Macos(\alpha)[/tex]
>>|
>>|[tex]a'(m+M+mcos^2(\alpha)) = (m+M)gsin(\alpha) + Macos(\alpha)[/tex]
>>|===========================================
Which eventually means I get:
>>|===========================================
>>|[tex]a' = (m+M)gsin(\alpha) + Macos(\alpha) / (m+M+mcos^2(\alpha))[/tex]
>>|===========================================
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
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