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Grade: 11
        
A light rod of length l is pivoted at distance l/4 from the left end has two masses Mo and m attached to its ends such that rod is in equilibrium . Find m in terms of Mo ?
11 months ago

Answers : (1)

Daniel dwayne
13 Points
							
LET THE LENGTH OF THE ROD BE (L) METRES
LET HEAVY MASS=M
AND LIGHTER MASS=m
PIVOTING THE ROD AT ¼ DISTANCE OF THE (L) FROM LEFT HAND SIDE
THUS MAKING THE LENGHT RATIO ¼.
NOW, HEAVIER MASS IS ATTACHED TO THE LEFT END OF THE ROD AND LIGHTER TO THE RIGHT END.
HERE THE DISTANCE OF THE HEAVIER MASS FROM THE PIVOT IS ¼ TIMES (L),
AND THE DISTANCE OF THE LIGHTER MASS FROM THE PIVOT IS ¾ TIMES (L)
NOW BALENCING THE MOMENT ABOUT THE POVOT,
M*1/4*L=m*(3/4*L)
m = M/3.
YOUR ANSWER IS THIS.
 
 
11 months ago
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