A spring balance inside an elevator is attache to a pulley with 1.5kg on one side and 3kg on other side.The elevator is going up with an accelaration of g/10.Find the reading of spring balance.
for 1.5 kg block T - mg - ma'[pseudo force] = ma T - 1.5g - 1.5g/10 = 1.5a mulitiply by 2 2T - 3g - 3g/10 = 3a ------------(1) for 3 kg block mg + ma' - T = ma 3g + 3g/10 -T= 3a ----------(2) (1) -(2) 2T - (-T) -3g-3g-3g/10-3g/10=0 3T -6g -6g/10 T -2g-2g/10=0 T = 22g/10
elevator is going upwards so a pseudo force acts on each masses..... and given that upward acceleration is g/10.... so for each masses we write equation as.... fist for 1.5 kg mass T-(mg+ma') =ma......where a' is given acceleration g/10 and ma' is pseudo force acting downward....... ......so putting values...T-(1.5g+1.5g/10) =1.5a..........(1)
now for other maas mg-T+ma' =ma..........................(2) now simplify these equations......u will get T and reading will be 2T/g and u will get answer 4.4kg..... and plzzzz rate me....if i helped u.....
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if i helped u plzzzzz rate me,,,,,,,
Good attempt ganesha and chimay, just one mistake by both of you. The reading in the spring balance will be T/g and not 2T/g. (Whenever you put a mass m on the balance another force of mg pulls it upward. That does not mean that reading is 2m.)
Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi
Moderation Team
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