Determine the reaction at points of constant 1,2 and 3. Assume smooth surfaces.

Given: The spheres are in equilibrium

To find: Reactions at points 1,2 and 3

#### Solution

Considering both the spheres as a single body

The system of two spheres is in equilibrium

Applying conditions of equilibrium:

ΣFy=0

R_{1}cos25 + R_{3}cos15 – g - 4g = 0

R_{1}cos25 + R_{3}cos15 = 5g ……(1)

ΣFx = 0

R_{1}sin25 - R_{3}sin15 = 0 …….(2)

Solving (1) and (2)

R_{1}= 19.75 N and R_{2} = 32.2493 N …….(3)

Let the reaction force between the wo spheres be R_{2} and it acts at an angle α with X-axis

Sphere A is in equilibrium

Applying conditions of equilibrium

ΣFy=0

R_{1}cos25 - R_{2}sinα – g = 0

R_{2}sinα = 8.0896 ……..(4) (From 3)

ΣFx=0

R_{1}sin25 - R^{2}cosα = 0

R_{2}cosα = 19.75sin25

R_{2}cosα = 8.3467 ………(5)

Squaring and adding (4) and (5)

R_{2}^{2}(cos^{2}α+sin^{2}α) = 135.1095

R_{2}=11.6237 N

Dividing (4) by (5)

`(R_2sinalpha)/(R_2cosalpha)=8.0896/8.3467`

α = tan-^{1}(0.9692)

=44.1038^{o}R_{1}=19.75 N (75o with positive direction of X-axis in first quadrant)

R_{2}=11.6237 N (44.1038o with negative direction of X-axis in third quadrant)

R_{3}=32.2493 N (75o with negative direction of X axis in second quadrant)