suryakanth AskiitiansExpert-IITB
Last Activity: 14 Years ago
Dear rameeja,
The force required to lift the train is
F = mg
F = 200*1000*9.81 = 1962000 N
consider angle theta = angle a
Fa+Fb cos(a) = F net
Fb sin(a) = mg = 1962000
Fb* (1/2) = 1962000
Fb = 1962000*2 = 3924000
Fb = 3924000
Fa+Fb cos(a) = F net
F net = 170000 + 3924000*(cos(30)) = 3568283.6844
and acceleration = F/m = 11.89
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
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Suryakanth –IITB