# Acceleration of 3 Blocks w/ Pulley

Coop

## 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.

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

$$\sum F_{x} = (T_{m_{2}})_{x} + (T_{m_{1}})_{left} + (T_{m_{1}})_{right}$$

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

So, $$\sum F_{x} = (T_{m_{1}})_{right}$$

And,

$$\sum F_{y} = (T_{m})_{y} + w_{y}$$

So,

$$F_{net} = \sum F_{x} + \sum F_{y} = (T_{m_{1}})_{right} + (T_{m})_{y} + w_{y}$$

But since,

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

So you are left with $$F_{net} = m_{net}*a = w_{y} = m*g$$

Then, you can solve

$$F_{net} = 4.9 kg*a = 1.5 kg*9.81\frac{m}{s^2}$$

$$a = 3.00 \frac{m}{s^2}$$

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

Thanks,
Coop

#### Attachments

• Practice MT1 #9.png
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