The Acceleration of a Block on an Incline Plane: Formula and Minimum Ratio

AI Thread Summary
The discussion focuses on calculating the acceleration of a block on an incline when a mass is hung from a pulley. The initial formula presented for acceleration was incorrect, as it did not account for the combined mass of both the block and the hanging mass. The correct expression for the minimum ratio of the hanging mass to the block mass was confirmed, emphasizing the importance of treating them as a single system. Participants acknowledged the value of peer review in identifying mistakes before submission. The conversation highlights the relevance of this problem in academic assessments.
tony873004
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Can someone tell me if I did this right?

An object of mass m is hung from a rope that passes over a pulley at the top of a ramp and is attached to a block of mass M.

(a) Assume m is large enough that the block accelerates up the ramp. Find an expression for the block’s acceleration.

(b) From your result for part (a), determine the minimum ratio m/M such that the block, once moving, accelerates up the ramp.


(the diagram is simple, M is on the incline plane and m is hanging straight down)

a = \frac{Force}{mass}

a=\frac{Mg sin\theta + \mu Mg cos \theta -mg}{M} (answer for part A)

Set the force equal to 0
Mg sin \theta + \mu Mg cos \theta - mg = 0
Mg sin \theta + \mu Mg cos \theta = mg

factor out Mg
Mg(sin \theta + \mu cos \theta) = mg

the g's cancel

m = M(sin \theta + \mu cos \theta)

m/M = sin \theta + \mu cos \theta
 
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AWKWARDLY,the answer to poin b) is correct (the minimum value is the one u found),but the acceleration you find is incorrect,assuming the rope is inextensible (the same tension in it in every point)...

How about correcting your mistake?

Daniel.
 
Thanks, Dex.
I'm guessing that I should have M+m in the denominator of my first answer instead of simply M. They must accelerate at the same rate so should be treated as a single mass.

By awkardly, did you mean because I had to get the first part to get the second part?
 
Yes,exactly,you got the point perfectly.Well done!

Daniel.
 
Thanks, Dextercioby. I'd have turned it in as is if you didn't make me go back and look!
 
Good thing that u decided to check by posting it here.

Daniel.
 
dextercioby said:
Good thing that u decided to check by posting it here.

Daniel.

You couldn't be more right. We had a mid term today, and this exact problem, but with specific numbers, was on the exam. I'd have missed it had we not gone through it last night!
 
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