Determining the Speed of a Cylinder in a Cable and Pulley System

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Homework Statement



The cylinder C is being lifted using the cable and pulley system shown. If point A on the cable is being drawn toward the drum with a speed of 2 m/s, determine the speed of the cylinder.

bfql2a.png


Homework Equations



[tex]2s_{A} + s_{b} = l[/tex]

The Attempt at a Solution



I set my points to this:

ftk2dd.png


I don't think its right because I am getting a negative number when it should be positive.

[tex]2s_{A} + s_{b} = l[/tex]

[tex]2v_{A} + v_{b} = 0[/tex]

[tex]v_{b} = -2(2m/s) = -4 m/s[/tex] This would mean that the cylinder is going down not up.
 
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KillerZ said:
The cylinder C is being lifted using the cable and pulley system shown. If point A on the cable is being drawn toward the drum with a speed of 2 m/s, determine the speed of the cylinder.
…
I don't think its right because I am getting a negative number when it should be positive.
…
This would mean that the cylinder is going down not up.

Hi KillerZ! :wink:

I think the cylinder does go down when the cable is drawn up.

But I don't think it's 2:1.

Try using sc instead of sb, where sc is the distance between the two lowest pulleys …

and use the fact that the total length of the string is constant. :smile:
 
So sc would be like this the difference between s1 and s2?

2yoocxz.png
 
I got it:

103ues4.png


[tex]s_{B} + (s_{b} - h) + (s_{B} - h - s_{A}) = l[/tex]

[tex]3s_{B} - s_{A} - 2h = l[/tex]

[tex]3v_{B} - v_{A} - 0 = 0[/tex]

[tex]v_{B} = -v_{A}/3 = -0.667m/s = 0.667m/s[/tex] up
 
Hi KillerZ! :smile:

Yes, except it's +vA/3.

(ignore what I said originally … I misread the diagram … the cylinder does go up! :redface:)