 #1
 40
 0
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
Attachments

11.1 KB Views: 392