Determine reactions force:
Three sphere A, B and C weighing 400N, and 200N and having radii 400mm, 600mm and 400mm respectively are placed in trench as shown in the figure given below. Treating all contact surfaces as smooth, determine reactions developed
.: From the figure given below
Sinα = BD/AB = (600 - 400)/(400 + 600) = 0.2
= 11.537° Referring to FBD of sphere A (Fig a)
R_{2}cosα = 200
R_{2} = 200/cos11.537° = 204.1 N .......ANS
And R_{1} - R_{2}sinα = 0
R_{1} = 40.8N .......ANS
Referring to the FBD of sphere C,
Sum of forces parallel to the inclined plane = 0
R_{4}cosα - 200cos45° = 0
R_{4} = 144.3 N .......ANS
Sum of forces perpendicular to the inclined plane = 0
R_{4}cos(45 -α ) - R_{3}cos45° = 0
R_{3} = 170.3N .......ANS
Refer to FBD of cylinder B (as shown in figure)
V = 0
R_{6}sin45° - 400 - R2cos - R4cos (45 + α ) = 0
R_{6}sin 45° = 400 + 204.1 cos11.537° + 144.3cos56.537°
R_{6} = 961.0 N .......ANS
H = 0
R_{5 } - R_{2}sin - R_{4}sin (45 +α ) - R_{6}cos45° = 0
R_{5} = 204.1 sin11.537 + 144.3sin56.537 + 961.0cos45°
R_{5} = 840.7 N .......ANS