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Unknown mass on a pulley system - Help!

  • #1

Homework Statement


The 100 kg block shown in the diagram takes 6.0s to reach the floor 1.0m below after being released from rest. What is the mass of the block on the left?


Homework Equations


Newton's Third Law
F=ma
F (a on b) = -F (b on a)


The Attempt at a Solution



I don't even know where to start, I was sick last class and can't figure out how to do this from the textbook. I've tried so far:
T1 = tension rope 1
T2 = tension rope 2
t0= 0s
t1= 6s
d=1.0m
g=9.8m/s^2
m1=unknown, find
m2=100kg
mass M: Fy = 0
T1-mg=0
T1=mg

T2= mg = (100kg)(9.8m/s^2)
= 980N

And now I'm lost.
 

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Answers and Replies

  • #2
Doc Al
Mentor
44,882
1,129
Start over.

(1) Calculate the acceleration from the given information.

(2) Apply Newton's 2nd law to each mass. This will allow you to solve for the tension in the rope and for the unknown mass.
 
  • #3
By using Newton's 2nd law,

d=0.5at2 (since v0 is zero),

I got an a=0.0556 m/s^2

Now, since they're on a pulley,

a= (m2g-m1g) / (m2+m1)

Plugging in, I get:

0.0556 = [(100kg*9.8m/s^2)-(m1*9.8m/s^2)] / 100kg + m1

rearranging, 5.56 + 0.0556 m1 = 980 - 9.8 m1
m1 = 100 kg

I know that's not the right answer, where am I going wrong?
 
  • #4
Doc Al
Mentor
44,882
1,129
The only error I see is here:
a= (m2g-m1g) / (m2+m1)
You have the signs of the two weights reversed. (Take positive to be in the direction of the acceleration.)
 

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