Piyush Kumar Maurya
Last Activity: 6 Years ago
First of all consider the circular face of the cylinder
Join the point A and B to the centre of circular face of Cylinder i.e O. These both will be equal to r( radius of circle).
Form a right ∆AOC inside the circle with side OA =r=1m, AC = a= 0.5m(given) . Similarly form a right angle ∆BOD inside the circle with side OB =r, BD = b=√3/2(given). Now we can see that the angle subtended by both these traingles at centre at different. Take Angle AOC = x and angle BOD = y.
Now in ∆AOC, . Sin x = P/H = a/r = 0.5/1 =1/2
So from here angle x = 30°
Now in ∆BOD, . Sin y = P/H = b/r = (√3/2)/1 = √3/2
So from here angle y= 60°
Reaction force on edge A is
F=mgcosx = 5*103 *9.8* cos 30° = 42.6kN
Similarly Reaction force on edge B is
F= mgcosy = 5*103 * 9.8* cos60° = 24.6kN
