## Determine reactions force, Mechanical Engineering

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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)

R2cosα  = 200

R2  = 200/cos11.537° = 204.1 N         .......ANS

And R1  - R2sinα    = 0

R1  = 40.8N                                               .......ANS

Referring to the FBD of sphere C,

Sum of forces parallel to the inclined plane = 0

R4cosα  - 200cos45° = 0

R4  = 144.3 N                                           .......ANS

Sum of forces perpendicular to the inclined plane = 0

R4cos(45 -α  ) - R3cos45° = 0

R3  = 170.3N                                             .......ANS

Refer to FBD of cylinder B (as shown in figure)

V = 0

R6sin45° - 400 - R2cos   - R4cos (45 + α   ) = 0

R6sin 45° = 400 + 204.1 cos11.537° + 144.3cos56.537°

R6 = 961.0 N                                           .......ANS

H = 0

R5  - R2sin   - R4sin (45 +α   ) - R6cos45° = 0

R5 = 204.1 sin11.537 + 144.3sin56.537 + 961.0cos45°

R5 = 840.7 N                                           .......ANS

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