Tension in a Two-Crate System on a Frictionless Surface

AI Thread Summary
The discussion revolves around calculating tension in a two-crate system on a frictionless surface, involving a 4 kg box and a 6 kg box. The initial equations set up for both boxes lead to confusion regarding the tension, with the first box yielding a tension of 10 N and the second suggesting zero tension. Participants clarify that the definitions of applied force (Fa) and net force (Fx) are crucial, emphasizing that Fa does not equal mass times acceleration (ma) but rather that the net force equals ma. There is also a noted error in the equations for the 6 kg box, as it should account for both the applied force and tension. The conversation highlights the importance of correctly identifying variables and ensuring equations are set up properly for accurate results.
babaypenguin88
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Homework Statement
Two crates, one with mass 4 kg and the other with mass 6 kg, sit on the frictionless surface of a frozen pond, connected by a light rope. A woman wearing golf shoes pulls horizontally on the 6 kg create with a force F that gives the crate an acceleration of 2.50n m/s^2 what is tension
Relevant Equations
F=ma
for 4kg box:
Fx = T
m1a=T
(4) (2.50) = T
10 = T

for 6 kg box: Fx = Fa - T
m2a = m2a - T
15 = 15 - T
15-15 = T
0 = T

I assumed the tension would be the same for both boxes, they are supposed to be, so what did I do wrong and is the answer 10 or zero?
 
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What are your definitions of Fx and Fa in both cases?
 
Orodruin said:
What are your definitions of Fx and Fa in both cases?
fa = force applied
fx = sum of forces in the x
 
babaypenguin88 said:
fa = force applied
fx = sum of forces in the x
You ought to have a diagram of the scenario. There are two boxes, hence two objects with forces on them.
 
babaypenguin88 said:
fa = force applied
fx = sum of forces in the x
Then Fa is not necessarily equal to ma. Only the net force is equal to ma.
 
babaypenguin88 said:
for 6 kg box: Fx = Fa - T
m2a = m2a - T
I think you have a typo in your 2nd equation. You can't have both ##F_x## and ##F_a## equal to ##m_2a##.
 
Further to @Mister T's remark, your equation for the 4kg block has only one unknown, T, whereas that for the 6kg block has/should have two, F and T. Consequently the second equation is only of interest if you wish to find F.
 
haruspex said:
Further to @Mister T's remark, your equation for the 4kg block has only one unknown, T, whereas that for the 6kg block has/should have two, F and T. Consequently the second equation is only of interest if you wish to find F.
ok thanks you
 
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