Why Are My Calculations for the Window Washer's Pulley System Incorrect?

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The discussion revolves around calculating the forces involved in a window washer's pulley system. To raise herself at a constant speed, the washer must exert a downward force equal to her weight, which is 686.7 N for a 70 kg mass. When increasing the pulling force by 8%, the net force must be recalculated to find the resulting acceleration. Participants emphasize the importance of understanding the relationship between the pulling force, tension in the rope, and the weight of the system. Clarifications are made that while the weight remains constant, the net force and acceleration can change based on the applied force.
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A window washer pulls herself upward using the bucket-pulley apparatus shown in Fig. 4-48.(heres the pic. http://img126.imageshack.us/i/physics.jpg/) The mass of the person plus the bucket is 70 kg.
(a) How hard must she pull downward to raise herself slowly at constant speed?
(b) If she increases this force by 8%, what will her acceleration be?
For some reason i got wrong answers 686.7 N and 10.59 m/s^2, Why?
 
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Don't just give your answers--show how you got them.

Hint: What's the relationship between how hard she pulls on the rope and the tension in the rope? How many times does the rope pull up on the "person+bucket" system?
 
Hi! I am actually working on a question that is pretty much identical, with different number for weight and % of increase. I was able to get part A.. F=m*a, but then you divide the answer by 2, because the force is split on either side of the pulley. I'm having trouble with B though. I tried to sub in the new force (answer+10.1%... or whatever % was in your question) and solve for acceleration using the same formula F=ma, but I'm still not getting the right answer. Please help!
 
PHYCCHICK said:
I'm having trouble with B though. I tried to sub in the new force (answer+10.1%... or whatever % was in your question) and solve for acceleration using the same formula F=ma, but I'm still not getting the right answer. Please help!
Don't forget that F stands for the net force and that gravity doesn't change.
 
I'm confused? If the acceleration remains the same (9.8) then how do I solve for a? And what other forces would be included in the net force?
 
PHYCCHICK said:
If the acceleration remains the same (9.8) then how do I solve for a?
The force of gravity--the weight--remains the same, not the acceleration!
And what other forces would be included in the net force?
Just the weight and the force from the ropes.
 
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