SOLUTION: I REALLY NEED HELP HERE.... A 3M BAR IS USE TO LIFT A STONE. A 90 KG PERSON EXERTS ALL HIS WEIGHT TO ONE END OF THE BAR WHICH IS 2 1/2M FROM THE FULCRUM AND BALANCES THE STONE. TH

Algebra ->  Inverses -> SOLUTION: I REALLY NEED HELP HERE.... A 3M BAR IS USE TO LIFT A STONE. A 90 KG PERSON EXERTS ALL HIS WEIGHT TO ONE END OF THE BAR WHICH IS 2 1/2M FROM THE FULCRUM AND BALANCES THE STONE. TH      Log On


   

Question 181321: I REALLY NEED HELP HERE....
A 3M BAR IS USE TO LIFT A STONE. A 90 KG PERSON EXERTS ALL HIS WEIGHT TO ONE END OF THE BAR WHICH IS 2 1/2M FROM THE FULCRUM AND BALANCES THE STONE. THE WEIGHT OF THE STONE IS APPROXIMATELY.
A. 500
B. 450
C. 400
D. 45KG
Found 2 solutions by vleith, Fombitz:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
The force at the end you are on is given by 90%2A+distance+to+the+fulcrum
90%2A+2.5
Since you balance, the force on the other end is X%2A+0.5
90%2A2.5+=+x+%2A+0.5
450+=+x

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Set up a moment (force x distance) equilibrium equation.
F%5B1%5Dd%5B1%5D=F%5B2%5Dd%5B2%5D
.
.
.
Find the associated distances from the pivot for both forces.
d%5B1%5D=2.5
d%5B1%5D%2Bd%5B2%5D=3
d%5B2%5D=0.5
.
.
.
The force, in this case, is gravity,
F=mg
m%5B1%5Dgd%5B1%5D=m%5B2%5Dgd%5B2%5D
m%5B1%5Dd%5B1%5D=m%5B2%5Dd%5B2%5D
90%282.5%29=m%5B2%5D%280.5%29
m%5B2%5D=90%282.5%2F0.5%29
m%5B2%5D=90%2A5=450
.
.
.
B.)450 kg.