Newton's laws,solving for tension (smthn wrong with my work)?

[JadeLove]

A flat-topped toy cart moves on frictionless wheels, pulled by a rope under tension The mass of the cart is m1. A load of mass m2 rests on top of the cart with the coefficient of static friction u between the cart and the load. The cart is pulled up a ramp that is inclined at an angle theta above the horizontal. The rope is parallel to the ramp. What is the maximum tension T that can be applied without causing to load to slip?

Okay. I figure this problem is fairly simple right?
(i) Fx = m2a
(ii) ff = m2a
(iii)Fy = 0
(iv)Fn - m2gcos = 0
(v) Fn = m2gcos
Plug v into (ii)
(vi) um2gcos = m2a
(vii) a = ugcos
Now sum forces for the cart
(vii) T - ff = (m1+m2)a
Plug in (vi) and (vii)
T - um2gcos = (m1+m2)(ugcos)
Here's my problem. The final answer is simply T = (m1+m2)(ugcos). Meaning there's an obvious problem with my work here...

What's the problem here? I'm so confused! My teacher suggested "adding m2gsin" to my fx equation, so like m2a + m2gsin? Because I tried that out and it still didn't work. Any help is great help, thanks in advance

 

[Doc Al]

(i) Fx = m2a
(ii) ff = m2a

What about gravity?

(iii)Fy = 0
(iv)Fn - m2gcos = 0
(v) Fn = m2gcos

OK.

Plug v into (ii)
(vi) um2gcos = m2a
(vii) a = ugcos

You'll need to fix this since you left out gravity on m2.

Now sum forces for the cart
(vii) T - ff = (m1+m2)a

The cart has mass m1, not m1 + m2. (And you forgot about gravity again.)

 

[JadeLove]

What about gravity?

OK.

You'll need to fix this since you left out gravity on m2.

The cart has mass m1, not m1 + m2. (And you forgot about gravity again.)

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Right, so does that mean it's m2a + m2gsin, with gravity added? or am I wrong??

 

[Doc Al]

Right, so does that mean it's m2a + m2gsin, with gravity added? or am I wrong??

Adding gravity to your m₂ equation means adding -m₂gsinθ to ƩFx.

 

[JadeLove]

Adding gravity to your m₂ equation means adding -m₂gsinθ to ƩFx.

Okay, thanks ^__^