- #1

Coop

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## Homework Statement

**There's a picture with the problem attached**

A block with mass m = 1.5 kg hangs to the right of two blocks with masses m2 = 2.2 kg and m1 = 1.2 kg which sit on a table. All three blocks are connected with a string and pulley. Neglecting the mass of the pulley, string and any friction, find the acceleration.

## Homework Equations

F = ma

## The Attempt at a Solution

The answer key says that m*g = (m + m1 + m2)*a

Is my reasoning below correct in coming to that conclusion? I was at first confused at why the tension forces are not shown.

I ignore the weight and normal force of the boxes on the table, since they cancel

[tex]\sum F_{x} = (T_{m_{2}})_{x} + (T_{m_{1}})_{left} + (T_{m_{1}})_{right}[/tex]

But [tex] (T_{m_{2}})_{x} + (T_{m_{1}})_{left} [/tex] cancel, right? Because they are equal but opposite?

So, [tex]\sum F_{x} = (T_{m_{1}})_{right}[/tex]

And,

[tex]\sum F_{y} = (T_{m})_{y} + w_{y}[/tex]

So,

[tex]F_{net} = \sum F_{x} + \sum F_{y} = (T_{m_{1}})_{right} + (T_{m})_{y} + w_{y}[/tex]

But since,

[tex](T_{m_{1}})_{right} + (T_{m})_{y}[/tex] are equal but opposite, they too cancel, right?

So you are left with [tex]F_{net} = m_{net}*a = w_{y} = m*g[/tex]

Then, you can solve

[tex]F_{net} = 4.9 kg*a = 1.5 kg*9.81\frac{m}{s^2}[/tex]

[tex]a = 3.00 \frac{m}{s^2}[/tex]

Does my reasoning make sense of how the tensions cancel and so are not factored into the equation?

Thanks,

Coop