Engineering Mechanics

(^40) „„„„„ A Textbook of Engineering Mechanics
Example 3.12. The lever ABC of a component of a machine is hinged at B, and is subjected
to a system of coplaner forces as shown in Fig. 3.19
Fig. 3.19.
Neglecting †friction, find the magnitude of the force (P) to keep the lever in equilibrium.
Also determine the magnitude and direction of the reaction at B.
Solution. Given : Vertical force at C = 200 N and horizontal force at C = 300 N.
Magnitude of the force (P)
Taking moments about the hinge B and equating the same,
P × 10 sin 60° = 200 × 12 cos 30° + 300 × 12 cos 60°
P × 10 × 0.866 = 200 × 12 × 0.866 + 300 × 12 × 0.5
8.66 P= 2078 + 1800 = 3878
∴
3878
447.8 N
8.66
P== Ans.
Magnitude of the reaction at B
Resolving the forces horizontally,
ΣH = 300 + P cos 20° = 300 + 447.8 × 0.9397 = 720.8 N
and now resolving the forces vertically,
ΣV = 200 – P sin 20° = 200 – 447.8 × 0.3420 = 46.85 N
∴ Magnitude of the reaction at B,
(^) RH V=Σ +Σ =( )^22 ( ) (720.8)^2 +(46.85)^2 =722.3 N Ans.
Direction of the reaction at B
Let θ = Angle, which the reaction at B makes with the horizontal.
∴
46.85
tan 0.0650
720.8
V
H
Σ
θ= = =
Σ
or θ = 3.7°Ans.
Note. Since both the values of ΣH and ΣV are +ve. therefore resultant lies between 0° and 90°.
3.14.COMPOUND LEVERS
A lever, which consists of a number of simple levers is known as a compound lever, as shown in
Fig. 3.20 (a) and (b).
† This point will be discussed in more details in the chapter on ‘Principles of Friction’