Tension in a Two-Crate System on a Frictionless Surface

[babaypenguin88]

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?

 

[Orodruin]

What are your definitions of Fx and Fa in both cases?

 

[babaypenguin88]

What are your definitions of Fx and Fa in both cases?

fa = force applied
fx = sum of forces in the x

 

[PeroK]

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.

 

[Orodruin]

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.

 

[Mister T]

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 $m_2a$.

 

[haruspex]

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.

 

[babaypenguin88]

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