find out the magnitude of the force to keep

Question

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_BH  = Resolved part of R_B along horizontal direction BE

R_BV = 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_BH  - P cos20°

∑H = 300 + R_BH  - 32.1cos20°                                                              ...(i)

∑v = 200 + R_BV  - Psin20°

∑v = 200 + R_BV  - 32.1sin20°                                                             ...(ii)

As the lever ABC is in equilibrium

∑H = RV  = 0, We get

R_BH = -269.85KN

R_BV = -189.021KN

	RB = {(RBH)2 +(RBV)2}1/2
	RB = {(-269.85)2 + (-189.02)2}1/2
	RB = 329.45KN	.......ANS

Let θ= Angle made by line of action of R_B with horizontal Then, tanθ )= R_BV/R_BH  = -189.021/-269.835

θ= 35.01°                                                                       .......ANS