
#1
Jan313, 01:03 PM

P: 35

1. The problem statement, all variables and given/known data
A man is trying to lift a box using three pulleys and two peices of rope. Each pulley can turn without friction and has negligible mass.The first rope, attached to the man, paases down under pulley A, up over pulley B and down to the frame of pulley C. The second rope is tied to the frame of pulley A and passes down under pulley C and up to the ceiling. If the box weighs 60N with what force must the man pull on the first rope to lift the box? 2. Relevant equations 3. The attempt at a solution To start off i know i have to draw an FBD. but would there be 3 FBDs for the pulleys or just one for the mass? 



#2
Jan313, 01:43 PM

P: 154

The ropes are wrapped around in such a way that just one FBD won't help you.




#3
Jan313, 01:45 PM

Mentor
P: 11,409

If you assume that the setup is in static equilibrium, then given that the pulleys and ropes are massless and frictionless you should be able to label the tensions in all the rope segments with a bit of careful thought. Start by assigning the unknown tension T to the rope held by the person (the one you've made red in your diagram). Can you determine the tension in the other rope?




#4
Jan313, 02:04 PM

P: 35

Pulleys and Mass
so would the tension be the same for both ropes and also when i am finding the tension im getting a value of zero?




#5
Jan313, 02:06 PM

P: 154





#6
Jan313, 02:08 PM

P: 35

ok so i would have an equation for each the ropes then....but when i am calculating tension i am getting zero...i thinmk my FBD equation is wrong. For the green rope would it be Fnet=mg + 3T




#7
Jan313, 02:10 PM

P: 154





#8
Jan313, 02:18 PM

P: 35

the net force or mass x acceleration and i think thats where i am wrong ..would it be 2T instead?




#9
Jan313, 02:20 PM

P: 154

What if you considered each pulley and found a relation between the tensions?




#10
Jan313, 02:22 PM

P: 35

so draw an FBD for each pulley and then try to find tension?
im assuming the tension in pulley A and B will be equal? 



#11
Jan313, 02:28 PM

P: 154

Yeah do that! And yes, the tensions in the ropes of both pulleys are same, since they are friction less.




#12
Jan313, 02:38 PM

P: 35

ok so heres what i got
for pulley A i have 2 tensions pulling up and one pulling down for B i got on tension pulling up and two pulling down i dunno if thats right and also how would i do C 



#13
Jan313, 02:41 PM

P: 154

I suggest you only focus on pulleys A and C. The FBD of pulley B is not necessary.
EDIT: If you can talk about the tensions for pulley A, why can't you do the same for C? Post some equations that you can write. 



#14
Jan313, 02:45 PM

P: 35

ok so pulley A will have 2 tensions up and one tension down
how would i do pulley C? would it also have 2 tensions pulling up and the mass pulling down? 



#15
Jan313, 02:46 PM

P: 154





#16
Jan313, 02:50 PM

P: 35

ok so i got 2 equations:
1. ma= T 2. ma = 2T  mg would i now add botht these equations together? 



#17
Jan313, 02:57 PM

P: 154

There are 2 different tensions in each of the ropes.
There is no acceleration because of what gneill said. Before you said that for pulley A, there were 2 tensions up and 1 down, and same for C. Label the tensions for each color and put them in equations and solve for both. Also, use 60 N instead of mg. 



#18
Jan313, 03:03 PM

P: 35

yeah but thats the thing for A i am getting the tension to be zero.
The equation i have for A is ma=t is this right . Also for pulley C i got 30N 


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