Find out the magnitude of the force to keep equilibrium:
The lever ABC of a component of machine is hinged at point B, and is subjected to a system of coplanar forces. Neglecting the friction force, find out the magnitude of the force P to keep the lever in the equilibrium.
Sol.: The lever ABC is in equilibrium under action of the forces 200KN, 300KN, P and RB, where RB needed reaction of the hinge B on the lever.
Thus the algebraic sum of the moments of above forces about any point in the plane is zero. Moment of RB and B is zero, as the line of action of RB passes through B.
Taking moment about point B, we obtain
200 × BE  300 × CE  P × BF = 0 as
CE = BD,
200 × BE  300 × BD  P × BF = 0
200 × BC cos30°  300 × BC sin30°  P × AB sin60° = 0
200 × 12 × cos30°  300 × 12 × sin30°  P × 10 × sin60° = 0
P = 32.10KN .......ANS
Let
R_{B}_{H} = Resolved part of R_{B} along horizontal direction BE
R_{B}_{V} = Resolved part of R_{B} along horizontal direction BD
∑H = Algebraic sum of Resolved parts of forces along horizontal direction
∑v = Algebraic sum of Resolved parts of forces along
vertical direction
∑H = 300 + R_{B}_{H}  P cos20°
∑H = 300 + R_{B}_{H }  32.1cos20° ...(i)
∑v = 200 + R_{B}_{V}  Psin20°
∑v = 200 + R_{B}_{V}  32.1sin20° ...(ii)
As the lever ABC is in equilibrium
∑H = RV = 0, We get
R_{B}_{H} = 269.85KN
R_{B}_{V} = 189.021KN

R_{B} = {(R_{BH})^{2} +(R_{BV})^{2}}^{1/2}


R_{B} = {(269.85)^{2} + (189.02)^{2}}^{1/2}


R_{B} = 329.45KN

.......ANS

Let θ= Angle made by line of action of R_{B }with horizontal Then, tanθ )= R_{BV}/R_{B}_{H} = 189.021/269.835
θ= 35.01° .......ANS