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Tension force

by physicsnub
Tags: force, tension
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physicsnub
#1
Nov30-03, 06:42 PM
P: 6
There is a pulley system with 2 masses, m1 = 3.2kg and m2 = 1.2 kg

the pulley is massless and frictionless.

What I have to do is determine the acceleration of the system and the tension force in the rope. How would I find the acceleration?
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sridhar_n
#2
Nov30-03, 06:59 PM
P: 85
I'll just give you a small hint. The Masses are suspended from the pulley on either side of the uplley. So the heavier mass is going to exert a force F on the lighter mass and will try to make it move towards itself. This force exerted by the heavier mass - tension will be equal to the force on the lighter mass. This force of the lighter mass will be equal will be equal to mass times the accln.

Can u find the accln now?


Sridhar
Doc Al
#3
Nov30-03, 06:59 PM
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Start like this: For each mass, draw a picture showing all the forces acting ON that mass. Then apply Newton's 2nd law (F=ma) to each.

AD
#4
Nov30-03, 07:02 PM
P: 73
Tension force

The larger mass will move downward and the smaller mass will move upward.

Just use the equations

m1g - T = m1a

and

T - m2g = m2a

To find the net force acting on each.
physicsnub
#5
Nov30-03, 07:22 PM
P: 6
Would the Acceleration be 12m/s^2, assuming gravity is 10m/s^2?

Originally posted by AD
The larger mass will move downward and the smaller mass will move upward.

Just use the equations

m1g - T = m1a

and

T - m2g = m2a

To find the net force acting on each.

I have not yet been introduced to those specific equations, I have just kind of been trying to figure this out on my own, could you explain them a little bit?
AD
#6
Nov30-03, 07:25 PM
P: 73
No. Think about it. How could the acceleration be greater than g?
physicsnub
#7
Nov30-03, 07:32 PM
P: 6
Youre right, I wasnt thinking about that right. What i'm thinking is if the 3.2 kg mass didnt have the 1.2 kg it would fall at 10m/s^2, right? but it has a 1.2 kg mass slowing it down a bit, so would i just take 10 - 1.2 kg for the acceleration?
sridhar_n
#8
Nov30-03, 07:36 PM
P: 85
No the accln is not and will not be greater than g:

This is because, the weight of both the bodies will act downwards while the accln on the lighte body due to the heavier body will act upwards. (Get the picture???) Hence the force eqtn wil be like this:

[tex]m_{1} * g - T = m_{2}*(g-a)[/tex]

So that the net accln of the lighter mass will not be greater than g.

Sridhar
Doc Al
#9
Nov30-03, 07:39 PM
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Originally posted by physicsnub
but it has a 1.2 kg mass slowing it down a bit, so would i just take 10 - 1.2 kg for the acceleration?
No. Do you understand the equations that AD laid out for you?

You are correct that if the mass were not connected to the rope, it would fall with acceleration = g. But it is connected to the rope, which pulls with tension T, thus reducing its acceleration. To find out what the acceleration is, don't cut corners... solve the equations!
AD
#10
Nov30-03, 07:42 PM
P: 73
I'll explain the equations. They're to determine the net force on each mass.

m1g - T = m1a

The net force on the larger mass, m1, is equal to its weight minus the tension in the rope.

The accleration of both masses is equal but the smaller one is travelling upward and the larger one is travelling downward.

For the second equation

T - m2g = m2a

The net force on the lighter mass is equal to the tension in the rope minus its weight, since the tension is greater than the weight because it is moving upwards.

What you should do is add these two equations and solve for 'a.'
AD
#11
Nov30-03, 07:59 PM
P: 73
?

Haven't you already posted that?
physicsnub
#12
Dec2-03, 05:55 PM
P: 6
Ok, I had to stop working on this problem, because I had to work on stuff that was actually assigned, I just started this physics class about a week ago, but I want to try this again :\.

The first equation m1g - T = m1a

for T do i just do T = (3.2kg)*(10m/s^2) ?
Would the first equation look like (3.2kg)*(10m/s^2) - 32 = (3.2kg)a ?
Doc Al
#13
Dec2-03, 06:59 PM
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Originally posted by physicsnub
The first equation m1g - T = m1a

for T do i just do T = (3.2kg)*(10m/s^2) ?
Would the first equation look like (3.2kg)*(10m/s^2) - 32 = (3.2kg)a ?
Go up to AD's last post in this thread. Read it very carefully. Do exactly what he says.
physicsnub
#14
Dec2-03, 07:12 PM
P: 6
well if thats not how i would go about starting that problem, i have no idea what hes telling me to do ive been trying to do it for a while now. i guess ill just wait until we get to this in class because this isnt really going anywhere, thanks though.
Sko
#15
Dec2-03, 08:28 PM
P: n/a
I'm trying to solve this problem too, and since it looks like he gave up I'll take over. I think I could do it but since I don't have my physics book home I'm a little confused about how to find the tension. is it [tex]m_1g + m_2g [/tex]
Doc Al
#16
Dec3-03, 04:36 AM
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I'm not sure what can usefully be added at this point (except just writing down the answer!). In a previous post, AD presented two equations. And he even told you how to solve them. Do it!
Sko
#17
Dec3-03, 03:46 PM
P: n/a
realized that was a needless question anyway I got

net acceleration= -4.5 m/[tex]s^2[/tex]
tension=17 N


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