How Do Pulley Systems Affect Tension Between Different Masses?

AI Thread Summary
The discussion focuses on calculating the tension in a pulley system involving three masses: m1 = 28 kg, m2 = 45 kg, and m3 = 12 kg. The initial attempt at solving for tension incorrectly assumed it was uniform throughout the cord, leading to confusion about the forces acting on the blocks. Participants emphasize that tension can vary along different segments of the cord, especially when masses are involved. A correct approach involves applying Newton's second law separately to each mass to account for the varying forces. Ultimately, understanding the relationship between tension and the gravitational forces acting on each block is crucial for solving the problem accurately.
dorkymichelle
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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)?

Homework Equations



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

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hi dorkymichelle! :smile:
dorkymichelle said:
… 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 :wink:
 
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...
 
hi dorkymichelle! :smile:
dorkymichelle said:
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

rubbish! :rolleyes:

if tension is bigger than gravity, then the block will accelerate up, won't it? :wink:
 
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