How to find tension of a single object in a pulley system?

[fissifizz]

Homework Statement

LETTER d only [/B]

Homework Equations

Fnet = ma

The Attempt at a Solution

So I considered 2m independently and set its net force equal to its gravity minus its tension.

(2m)a = Fg - Ft = (2m)g - Ft

I then rearranged the equation so that Ft = (2m)g - (2m)a = 2m(g-a)

There's no answer key, so I'm not sure if this is correct. I appreciate any assistance with this problem! Thanks!

 

[Herman Trivilino]

The problem is that your answer contains a. The directions say to express your answers in terms of m and g.

So, you need to find an expression for a in terms of m and g.

 

[fissifizz]

The problem is that your answer contains a. The directions say to express your answers in terms of m and g.

So, you need to find an expression for a in terms of m and g.

Ah, ok. Since all the blocks are moving in a system, I'm going to assume that the acceleration calculated in letter b will be used.

Fnet = (5m)a = 5mg - 4mg (I hope I'm calculating this correctly)
(5m)a = mg
a = g/5

So time to plug back in: Ft = 2m(g-a) = 2m(g-(g/5))
Ft = 2m(4g/5)
Ft = 8mg/5

It seems like that's it. Was my original methodology correct though?

 

[ehild]

Ah, ok. Since all the blocks are moving in a system, I'm going to assume that the acceleration calculated in letter b will be used.

What do you mean? The blocks are connected, and the magnitude of their acceleration is the same.

Fnet = (5m)a = 5mg - 4mg (I hope I'm calculating this correctly)

The net force is correct, mg, but the whole mass is not 5m.