(^62) „„„„„ A Textbook of Engineering Mechanics
Solution. Given : Weight of cylinder = 100 N
Fig. 5.12.
Let RA= Reaction at A, and
RB= Reaction at B.
The smooth cylinder lying in the groove is shown in Fig. 5.12 (a). In order to keep the system
in equilibrium, three forces i.e. RA, RB and weight of cylinder (100 N) must pass through the centre of
the cylinder. Moreover, as there is no *friction, the reactions RA and RB must be normal to the surfaces
as shown in Fig. 5.12 (a). The system of forces is shown in Fig. 5.12 (b).
Applying Lami’s equation, at O,
100
sin (180 – 40 ) sin (180 – 15 ) sin (15 40 )
RRAB==
°° °° °+°
or
100
sin 40 sin 15 sin 55
RRAB==
°°°
∴
100 sin 40 100 0.6428
78.5 N
sin 55 0.8192
RA
×° ×

°
Ans.
and
100 sin 15 100 0.2588
31.6 N
sin 55 0.8192
RB
×° ×

°
Ans.
Example 5.6. Two cylinders P and Q rest in a channel as shown in Fig. 5.13.
Fig. 5.13.
The cylinder P has diameter of 100 mm and weighs 200 N, whereas the cylinder Q has diameter
of 180 mm and weighs 500 N.

• This point will be discussed in more details in the chapter of Principles of Friction.