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Physics Strings and pulley

mathmaniac

Active member
Mar 4, 2013
188
The system is in equlibrium,find T1 and explain.
Thanks.
 

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Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
When I click on the link, it says the attachment is invalid. Could you please try to upload that again? Also, please post any work on the problem that you have done (I realize you may have had some work in the attachment, but of course, I can't see that yet.)
 

mathmaniac

Active member
Mar 4, 2013
188
Modified it.

My work?
I can see nothing is moving,so I conclude
4g+6g=T1+T
where g is acceleration due to gravity

I don't see anyway to go further.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Modified it.

My work?
I can see nothing is moving,so I conclude
4g+6g=T1+T
where g is acceleration due to gravity

I don't see anyway to go further.
On the sketch I'm going to call m1 = 4 kg and m2 = 6 kg. I'm also going to deal with the two masses as separate systems. I am defining upward as +y in both diagrams.

For m1:
Using Newton's 2nd I get T1 - T - m1*g = 0, so we get an equation for T1 in terms of T.

For m2:
Using Newton's 2nd I get -m2*g + T = 0.

Solve the m2 equation for T and plug it into the m1 equation.

-Dan
 

mathmaniac

Active member
Mar 4, 2013
188
T=6g and T1=6g+4g=10g
But this is not the answer I had been taught,it is 2g.

For m1:
Using Newton's 2nd I get T1 - T - m1*g = 0, so we get an equation for T1 in terms of T.
For m2:
Using Newton's 2nd I get -m2*g + T = 0
how can it be when some other forces like 4g and T1 are pulling T?
Same question for m1.

I'm also going to deal with the two masses as separate systems.
Things on the one side of the pulley have no effects on the other?
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
For m1:
Using Newton's 2nd I get T1 - T - m1*g = 0, so we get an equation for T1 in terms of T.
I got my T1's and T's reversed. This should read T - T1 - m1*g = 0

Sorry about that.

T=6g and T1=6g+4g=10g
But this is not the answer I had been taught,it is 2g.
Yes, that's the correct answer.

Things on the one side of the pulley have no effects on the other?
They do correlate...through the tension T.

If you wish to consider the system as one unit, go ahead (though you won't get an equation for T this way.) The trick is that the pulley "unbends" the problem into a straight line. +y upward on one side of the pulley implies +y being downward on the other. It can certainly be done, but requires a bit more work is all. Oooh! I almost forgot to say...any solution here requires that there is a tension T pulling upward on m1 and the same value for the tension is pulling upward on m2. If you aren't using ideal strings this goes out the window. When in doubt, assume an ideal string.

-Dan
 

mathmaniac

Active member
Mar 4, 2013
188
Please demonstrate how it would be like when the string is laid out in a straight line.
Thanks
Regards
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Please demonstrate how it would be like when the string is laid out in a straight line.
Thanks
Regards
"Ideal" pulleys do nothing more than change the direction of the tension. You can straighten them out but note when you do that the weight of the mass on the left is to the left, and the weight of the mass on the right is to the right. It is for this reason I prefer not to teach my classes about it...the concepts can get a bit confusing when all the forces get laid down into one line.

Using your problem as an example, let's define +y to the right. Then we have that T1 is a force acting to the left (so we would say it's in the negative direction), there is a weight acting in the negative direction, there is a tension acting in the positive direction, etc. It's certainly not an impossible calculation to do, but labeling and finding out what forces act in what direction is a bit too complicated for my taste.

-Dan