A wheel of radius 40cm rests against a step of height 20 cm. If the mass of the wheel is 2kg, then find the minimum horizontal force required to be applied perpendicular to axle, so as to make the wheel climb the step. Try this one on for size! This appeared in my school paper, and you won't find this in any book, as this was created by a teacher of my school who is an ex-iitian. Nobody has got it correct till now.
Asked by Arnabh Kumar | 11th Mar, 2011, 01:19: AM
We know that both weight of the wheel and normal force of the ground act through the center of wheel. Force of the step = fx.
Now, the wheel will remain stationary as long as Σtorque =0, i.e.
MgRsinΘ - NRsinΘ - FRcosΘ = 0, when N=0, however, the wheel will start loosing contact with
the ground and go over the step. This occurs when, F = MgtanΘ
R( 1-cosΘ) =h, the height of the step,
so, Θ = cos-1 (1-h/R) = cos-1 (1-20/40)
Then, F = 2X 9.8X tan((cos-1 (1-20/40))
F =33.9472 N
Is this right?
Hope this was useful.
Answered by | 10th Mar, 2011, 10:19: PM
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