Two Weights P=5kg & Q=7Kg are suspended from the end of a beam AB 5 metre long. These force are balanced by a force F acting at C in between AB. what is the magnitude of F & position of ‘C’
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Answer:
its easy
Explanation:
joMass 5kg will go down.
5g — T= 5a( T is the tension, a is the accln.) (1)
T--3g = 3a. (mass3kg will go up with accln. a) ( ,2)
T will be the same on both sides,pulley being frictionless, and string being inelastic.
Adding ( 2) and (1)
2g= 8a
a= 20 / 8 m/ s^2
From 1
50 —T = 5*20/8
T= (400 — 100)/8
T = 300 /8.Newton
Total pull or load on the pulley
2T= 600/8 Newton.= 75Newton
= 75 Newton.
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