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Please solve the problem in the image urgently... 
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Asked by subhrajayanta64 | 04 Sep, 2019, 11:31: PM
answered-by-expert Expert Answer
mass of the wheel is not given. Let us assume mass is m kg
 
It is assumed,  required pulling force P is meant for pulling up the wheel to the platform of height 150 mm from the base of the wheel.
 
change in potential energy  PE for the wheel to be pulled up by 150 mm is given by
 
PE  = mg×150×10-3 J   ............................(1)
 
While pulling up, radius OA has rotated by angle θ . If   Τ is the torque, then workdone W = Τθ  J ....................(2)
 
Torque Τ = P × OA = P × 300 × 10-3 N-m
 
Rotation angle θ = π/3 rad
( see the figure. OA initially at 30º with horizontal . Hence to make OA vertical, rotation angle is 60° )


 
By equating eqn.(1) and eqn.(2) and using the substitutions for Τ and θ , we get
 
mg ×150×10-3 = P × 300 × 10-3 × (π/3)   or     P  = 0.48×mg
 
 
Answered by Thiyagarajan K | 05 Sep, 2019, 11:14: AM

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