How Do Pulley Systems Affect Tension Between Different Masses?

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Homework Help Overview

The problem involves a system of three masses connected by cords over a pulley, with specific masses given. The original poster seeks to determine the tension in the cord connecting two of the masses and the distance one mass moves after a set time, while grappling with the dynamics of the system.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply Newton's second law to find the acceleration and tension in the system but encounters difficulties with the tension values. Some participants question the assumptions about tension being equal throughout the system and suggest that the presence of mass affects the tension differently.

Discussion Status

Participants are actively engaging with the problem, with some providing alternative perspectives on the tension in the system. There is a recognition of potential errors in the original poster's calculations, and discussions are ongoing regarding the correct application of forces and the implications of mass on tension.

Contextual Notes

There is a mention of a figure that is referenced but not provided in the thread, which may be critical for visualizing the problem setup. Additionally, participants are navigating the complexities of tension in a system with multiple masses and the effects of gravity.

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

  • fig_5_E.gif
    fig_5_E.gif
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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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