Incline Plane with Friction with one object hanging?


by myelevatorbeat
Tags: friction, incline, object, plane
myelevatorbeat
myelevatorbeat is offline
#1
Sep9-07, 06:16 PM
P: 55
1. The problem statement, all variables and given/known data

Find the acceleration reached by each of the two objects shown in Figure P4.49 if the coefficient of kinetic friction between the 7.00kg object and the plane is 0.250

Here is a picture of the problem: http://a496.ac-images.myspacecdn.com...bcabca499f.jpg

2. Relevant equations
F=ma


3. The attempt at a solution

I'm calling the 7.00 kg object M1 and the 12.0kg object M2.

<b>For M1</b>
(Fnet)x=ma
T+mgsin37degrees-fk=m1a

(Fnet)y=ma (a=o)
n1-mgcos37degrees=0
n1=mgcos37degrees


<b>For M2</b>
(Fnet)y=m2a
T-m2g=m2a
T=m2a+m2g

I take this equation and put it in the (Fnet)x equation I got for M1:
T+mgsin37degrees-fk=m1a
m2a+m2g+mgsin37degrees-fk=m1a
m2a+m2g+mgsin37degrees-(0.250)mgcos37degrees=m1a
m2g+mgsin37degrees-(0.250)mgcos37degrees=m1a-m2a
a=m2g+mgsin37degrees-(0.250)mgcos37degrees / (m1-m2)
a=7.64 m/s^s

Now, my question is: It says "Find the acceleration reached by <b>each</b> of the two objects. Now, I know with an ideal pulley the tensions on both side of the equation are equal, but I'm not sure if the acceleration is. I would assume so cause it's one rope and one rope can't move at two different speeds at once, but I could be wrong. So, is that the answer to the acceleration of both blocks or merely the acceleration of the 7.00kg object and there is further work to be done to find the acceleration of the 12.0kg object?
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learningphysics
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#2
Sep9-07, 06:23 PM
HW Helper
P: 4,125
You can assume accelerations are the same... But careful about signs... if you assume the acceleration of M1 is a acting upward to the right... then you must assume that the acceleration of M2 is a downward... so this will affect the signs in your equations.

also here:

<b>For M1</b>
(Fnet)x=ma
T+mgsin37degrees-fk=m1a
I think you should have T - m1gsin37 - fk = m1a. (minus instead of plus)

And since you've assumed that a is upward to the right for M1... it is downward for M2... so

T - m2g = m2(-a)
Doc Al
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#3
Sep9-07, 06:24 PM
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Quote Quote by myelevatorbeat View Post
<b>For M1</b>
(Fnet)x=ma
T+mgsin37degrees-fk=m1a
The tension and the component of the weight act in different directions.

In calling the acceleration "a", be sure to use a consistent sign convention for M1 & M2: If M1 accelerates up the incline, M2 must accelerate down.
Now, I know with an ideal pulley the tensions on both side of the equation are equal, but I'm not sure if the acceleration is. I would assume so cause it's one rope and one rope can't move at two different speeds at once, but I could be wrong.
That is correct and is essential to solving the problem.


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