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
RBH = Resolved part of RB along horizontal direction BE
RBV = Resolved part of RB 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 + RBH - P cos20°
∑H = 300 + RBH - 32.1cos20° ...(i)
∑v = 200 + RBV - Psin20°
∑v = 200 + RBV - 32.1sin20° ...(ii)
As the lever ABC is in equilibrium
∑H = RV = 0, We get
RBH = -269.85KN
RBV = -189.021KN
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RB = {(RBH)2 +(RBV)2}1/2
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RB = {(-269.85)2 + (-189.02)2}1/2
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RB = 329.45KN
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.......ANS
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Let θ= Angle made by line of action of RB with horizontal Then, tanθ )= RBV/RBH = -189.021/-269.835
θ= 35.01° .......ANS