Question

The lever ABC of a component of a machine is hinged at B, and is subjected to a system of coplanar forces. Neglecting friction, find the magnitude of the force P to keep the lever in equilibrium.

Answer 1

The lever ABC is in equilibrium under the action of the forces 200KN, 300KN, P and RB, where RB required reaction of the hinge B on the lever.

Hence the algebraic sum of the moments of above forces about any point in their plane is zero. Moment of RB and B is zero, because the line of action of RB passes through B. Taking moment about B, we get

200 × BE – 300 × CE – P × BF = 0 

since CE = BD, 

200 × BE – 300 × BD – P × BF = 0

200 × BC cos 30° – 300 × BC sin 30° – P × AB sin 60° = 0

200 × 12 × cos 30° – 300 × 12 × sin 30° – P × 10 × sin 60° = 0

P = 32.10 KN

Let 

R_BH = Resolved part of R_B along a horizontal direction BE 

R_BV = Resolved part of R_B along a horizontal direction BD 

∑H = Algebraic sum of the Resolved parts of the forces along horizontal direction 

∑v = Algebraic sum of the Resolved parts of the forces along vertical direction

∑H = 300 + R_BH – P cos 20°

∑H = 300 + R_BH – 32.1 cos 20° .......(i)

∑v = 200 + R_BV – Psin 20°

∑v = 200 + R_BV – 32.1 sin 20° ......(ii)

Since the lever ABC is in equilibrium

∑H = R_V = 0, We get

R_BH = –269.85KN

R_BV = –189.021KN

R_B = {(RBH)² +(RBV)²}^1/2

R_B = {(–269.85)² + (–189.02)2}^1/2

R_B = 329.45KN

Let θ = Angle made by the line of action of R_B with the horizontal 

Then, tanθ = R_BV/R_BH = –189.021/–269.835 

θ = 35.01°