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Posted on: 10:47:02 AM on August 5, 2019
Posted By:
TAMSIL GAUSIYA
The tricycle weighing 20 kg has a small wheel symmetrically placed 1 m behind the two large wheels, which are also 1 m apart. If the centre of gravity of machine be at a horizontal distance of 25 cm behind the front wheels and the rider whose weight is 40 kg, be 10 cm behind the front wheels. The thrust on each front wheel is


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Posted on: 12:16:05 PM on March 23, 2020
Solution : a) In the given figure, the tricycle with two front wheels A and B with one rear wheel C with center of gravity of machine G1 and that of rider G2. Now, force at G1=20×10=200N Force at G2=400 N Let FC = Force on ground under rear wheel C Force at D = FD=2FA acts on mid-point of AB. Taking torque about D, for equilibrium, Fc×100=200×25+400×10=9000 ∴ Thrust on each front wheel is N. ∴ 2N + F_(c)=Totalweight=600NthereforeN=600−FC2= 255 N





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