**Determine reactions force:**

**Thr****e****e 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 - *R*2cos - *R*4cos (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**