# Forces and pulley systems

## Homework Statement

In the figure here three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are m1 = 28 kg, m2 = 45 kg, and m3 = 12 kg. When the assembly is released from rest, (a) what is the tension in the cord connecting 2 and 3, and (b) how far does 1 move in the first 0.300 s (assuming it does not reach the pulley)?

F= ma
d= vit+1/2at^2

## The Attempt at a Solution

first I got Ft = m1a
Ft - (m2g+m3g) = (m2+m3)a
subtracted those 2 equations and got
-(m2g+m3g)=-(m2+m3)a-m1a
solved for a,
a = (m2g+m3g)/(m2+m3+m1)
a = 6.57176 m/s^2
I did b correctly using the kinematics equation but didn't get A correct.
I thought tension in string is equal for all parts so i did Ft = ma = 184.0 N, but that's wrong.
Is there a connection between block 2 and 3 that I'm missing?

The figure is in the attatchments

#### Attachments

• fig_5_E.gif
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tiny-tim
Homework Helper
hi dorkymichelle! … first I got Ft = m1a
Ft - (m2g+m3g) = (m2+m3)a
subtracted those 2 equations and got
-(m2g+m3g)=-(m2+m3)a-m1a
solved for a,
a = (m2g+m3g)/(m2+m3+m1)

you got a minus sign wrong near the start, so i'm not sure how you got the right result!
I thought tension in string is equal for all parts …

no, the tension is only the same along a continuous piece of string, not one with a mass in the way (or a frictional surface) …

do F = ma on the lowest block Hmm.. I did F=ma
F = (12)(6.57176) = 78.86 N
but that's the total forces which includes both tension and gravity. Since the block is not in freefall, then tension is bigger than gravity
Ft - mg = 78.86N
Ft-(12)(9.8)=78.86N
Ft-117.6=78.86N
Ft = 196.46 N
which is wrong...

tiny-tim
hi dorkymichelle! rubbish! if tension is bigger than gravity, then the block will accelerate up, won't it? 