Question

Three sphere A, B and C weighing 200N, 400N and 200N respectively and having radii 400mm, 600mm and 400mm respectively are placed in a trench as shown in fig (a), Treating all contact surfaces as smooth, determine the reactions developed.

Answer 1

From the fig (a)

Sinα = BD/AB = (600 - 400)/(400 + 600) = 0.2

α = 11.537°

Referring to FBD of sphere A Fig (a)

R₂ cosα = 200

R₂ = 200/cos 11.537° 

= 204.1 N

And R₁ – R₂ sinα = 0

R₁ = 40.8N

Referring to the FBD of sphere C [Fig (b)],

Sum of forces parallel to inclined plane = 0

R₄ cosα – 200 cos 45° = 0

R₄ = 144.3 N

Sum of forces perpendicular to inclined plane = 0

R₄ cos(45 – α) – R₃ cos 45° = 0

R₃ = 170.3N

Referring to FBD of cylinder B (Fig.(b)] 

∑V = 0

R₆ sin 45° – 400 – R₂ cosα – R₄ cos (45 + α) = 0

R₆ sin 45° = 400 + 204.1 cos 11.537° + 144.3 cos 56.537°

R₆ = 961.0 N

∑H = 0

R₅ – R₂ sin α – R₄sin (45 + α) – R₆ cos 45° = 0

R₅ = 204.1 sin 11.537 + 144.3 sin 56.537 + 961.0 cos 45°

R₅ = 840.7 N.