Two spheres A and B of weight 1000N and 750N respectively are kept as shown in the figure..Determine reaction at all contact points 1,2,3 and 4. Radius of A is 400 mm and radius of B is 300 mm

**Given ** : Two spheres are in equilibrium

W_{1}=1000 N

W_{2}=750 N

r_{A}=400 mm

r_{B}=300 mm **To find : Reaction forces at contact points 1,2,3 and 4**

#### Solution

BC = BP = 300mm = 0.3m

AP = 400 mm = 0.4 m

AB = AP + BP

= 0.7m

CO = BC + BO

0.7 = 0.3 + BO**BO = 0.4m**

In △AOB

`cos α = (BO)/(AB) = (0.4)/(0.7)`

α = cos^{-1}(0.5714) **α = 55.1501 ^{o }**

^{}

Forces R_{3},R_{4} and 1000N are under equilibrium at point A

**Applying Lami’s theorem**

`(R3)/(sin120) = 1000/(sin (150−alpha)) = (R4)/(sin (90+alpha))`

`(R3)/(sin120) = 1000/(sin (150−55.1501)) = (R4)/(sin(90+55.1501))`

Solving the equations**R _{3} = 869.1373 N**

**R**

_{4}= 573.4819 N

Forces R_{1},R_{2},R_{3} and 750N are under equilibrium at B

Applying conditions of equilibrium

**ΣF _{Y}=0 **

-R

_{3}sin α-750+R2=0

R

_{2}=869.1373 sin55.1501+750

R

_{2}=1463.2591 N (Acting upwards)

Applying conditions of equilibrium

**ΣF _{X}=0**

R

_{1}-R

_{3}cosα=0

R

_{1}=869.1373cos55.1501

**R _{1}=496.65 N(Acting towards right)**

Sr.no. |
Point |
Force |

1. |
R_{1} |
496.65 N(Towards right) |

2. |
R_{2} |
1463.2591 N(Towards up) |

3. |
R3 |
869.1373 N(55.1501^{o} in first quadrant) |

4. |
R4 |
573.4819 N(30^{o} in second quadrant) |