Why is tension (T) only added to one side of an Atwood Machine?

[PhysicsCanuck]

Homework Statement

Suppose in the same Atwood setup another string is attached to the bottom of m1 and a constant force f is applied, retarding the upward motion of m1.

If m1 = 5.00 kg and m2 = 10.00 kg, what value of f will reduce the acceleration of the system by 50%?

Relevant Equations

T = m1*a1 + m1*g
T = m2*a2 + m2*g

I solved for a1 prior to the force (f) being added.

-a1 = a2

and

T = m1*a1 + m1*g
T = m2*a2 + m2*g <--substitute -a1 = a2, multiply everything by -1, add the two equations in order to solve for a1 (and thus also a2)

T = m1*a1 + m1*g
-T = m2*a1 - m2*g

0 = m1*a1 + m1*g + m2*a1 - m2*g <-- substitute known values (m1 = 5.00kg, m2 = 10.00kg, g=9.8m/s^2), solve for a1
a1 = +3.266 m/s^2 (and thus a2 = -3.266 m/s^2)

The question then states a 50% reduction, so 3.266/2 = 1.633 m/s^2 for the following equations.
After grinding through this for hours (and knowing the final solution to be 24.5N), I was able to determine the following:

T' = m1*a1 + m1*g + F
T' = m2*a2 + m2*g

Since we know m1, m2, a1, a2, and g, we can solve for F, where F = 24.5NMy question is as follows: Since a force (and thus a tension) is being added to m1 (and thereby increasing the tension experienced in the line/rope/cable, wouldn't I also add that same force value to m2 since it is also experiencing more tension as they are connected and inextensible.

Can anyone explain this and help me better understand?

Thank you very much.

 

[BvU]

Hello Canuck, $\qquad$ $\qquad$ !

My question is as follows: Since a force (and thus a tension) is being added to m1 (and thereby increasing the tension experienced in the line/rope/cable, wouldn't I also add that same force value to m2 since it is also experiencing more tension as they are connected and inextensible.

Seems so sensible, doesn't it ?
But the extra F is really applied on one side only.
Compare with a see-saw: the balance shifts if you add an extra weight on one side only. And it shifts differently if you instead add it on the other side. Same with your Atwood machine.

'Try' it out for yourself with a simple example, e.g. when m₁ = m₂

(i.e. a given F on m₁ side versus same F on m₂ side )

 

[Orodruin]

The equations correspond to the force balance on the masses. The force is only acting on one mass. The change in the total force on the second mass is due to the change in the tension, which will be different in the two cases.

 

[PhysicsCanuck]

Hello Canuck, $\qquad$ $\qquad$ !

Seems so sensible, doesn't it ?
But the extra F is really applied on one side only.
Compare with a see-saw: the balance shifts if you add an extra weight on one side only. And it shifts differently if you instead add it on the other side. Same with your Atwood machine.

'Try' it out for yourself with a simple example, e.g. when m₁ = m₂

(i.e. a given F on m₁ side versus same F on m₂ side )

Thank you very much.
This type of question is novel for me, so I guess it will become clearer with time and more practiced examples.
Cheers