Modified Atwood's Machine on an Inclined Plane

  • Thread starter Labrack
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1. The problem statement, all variables and given/known data
Block B with a mass of (.75kg) is on an inclined plane attatched by a pulley to Block A which has a mass of (1.5kg). The angle given is 40 degrees. The coefficient of friction of Block B on the ramp is .29. What is the acceleration?


2. Relevant equations
Fnet=m*a
f=Fnormal*Coefficient of f


3. The attempt at a solution
My solution was Fnet = m*a
Fnet = Fapplied - f
Fapplied = the weight of Block A (m*g) = 14.7N
f = fx of Block B which equals sin(40)*7.36 = 2.25
Fnet = 12.5
therefore a=Fnet / m
m = mass of the system which is 1.5 +.75 = 2.25kg
a = 12.5 / 2.25
a= 5.56m/s/s


I don't think that this is correct. Our Physics teacher has showed us inclined plane problems with a = 0 and without Atwood's machine. He has also showed us an Atwood's Machine problem without an angle. So now I am stuck and I don't know how to accuratley solve this problem. Thanks for the help.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 

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Last edited:

Doc Al

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You left out the friction force.

Rather than doing the problem in a single step, you might want to apply Newton's 2nd law to each mass separately. You'll get two equations that you can solve together to find the acceleration. This method will allow you to solve all sorts of complicated problems.
 
Where do I need to input the frictional force?

I thought that the frictional force was the 'x' component of the weight of Block B which is 2.25N

@Doc Al, did you see the diagram I attatched? Thank you for looking at this :)
 

Doc Al

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Where do I need to input the frictional force?
It's a force acting on block B.

I thought that the frictional force was the 'x' component of the weight of Block B which is 2.25N
No, the x-component of the weight is another force on block B. (Hint: To find the friction force, first find the normal force between block and incline.)
 

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