Pulley dynamics problem 12-205 Hibbeler

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Homework Statement


I have posted the snapshot.

Homework Equations


I have written the distances from the datum line. Since we have two threads, I got two
equations.
[tex]2S_A +S_C=L_1[/tex]

[tex](S_B -S_C)+(h-S_C)=L_2[/tex]

where L1 and L2 are the lengths of the strings excluding the
parts which remain constant in time.

The Attempt at a Solution

I can now relate B and A as

[tex]\dot{S_A}=-\frac{\dot{S_B}}{4}[/tex]

[tex]\ddot{S_A}=-\frac{\ddot{S_B}}{4}[/tex]

So I get [itex]\dot{S_A} =-1[/itex] , which means the block A is going upwards.
Now the problem says that the speed of the cable being pulled at B is decreasing
at the rate of 2 ft/s2. So that means [itex]\ddot{S_B}=-2 ft/s^2[/itex].
So I get [itex]\ddot{S_A}= 0.5[/itex]. I got the first answer right. I have question about
the interpretation of the second answer. Since [itex]\ddot{S_A}[/itex] is positive, does
it mean the speed of block A is increasing ?. My second numerical answer is correct
though.
 

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hi IssacNewton! :smile:
IssacNewton said:
… So I get [itex]\dot{S_A} =-1[/itex] , which means the block A is going upwards.

Now the problem says that the speed of the cable being pulled at B is decreasing
at the rate of 2 ft/s2. So that means [itex]\ddot{S_B}=-2 ft/s^2[/itex].
So I get [itex]\ddot{S_A}= 0.5[/itex] …

Since [itex]\ddot{S_A}[/itex] is positive, does it mean the speed of block A is increasing ?

A's velocity is -1 downward, ie 1 upward.

A's acceleration is positive downward, ie negative upward, so the speed of 1 upward is decreasing. :wink:
 
tim, makes perfect sense. these pulley problems sometime throw me...