Find the resultant of the force acting on the bell crank lever shown. Also locate its position with respect to hinge B.

Given : Forces on the bell crank lever

To find : Resultant and it’s position w.r.t hinge B

#### Solution

Let the resultant of the system of forces be R and it is inclined at an angle θ to the horizontal The hinge is in equilibrium Taking direction of forces towards right as positive and towards upwards as positive

Applying the conditions of equilibrium

ΣFx = 0

Rx = 50cos60 + 120

= 145 N

Ry = -50sin60 -100

= -143.3013

R= `sqrt(R2/x+R2/y)`

`=sqrt(145^2+(-143.3013))^2`

= 203.8633 N

θ = tan-^{1 }`(Ry)/(Rx)`

`=tan^-1((143.3013)/145)`

=44.6624^{o}

Let the resultant force R be acting at a point x from the point A and it is at a perpendicular distance of d from point A

Taking moment of forces about point A and anticlockwise moment as positive

Applying Varigon’s theorem,

203.8633 x d = -(100 x 20) – (120 x 40cos30)

d = -30.2012 cm = 30.2012 cm ………..(as distance is always positive)

sin 44.6624 =`x/30.2012`

x = 21.2293 cm

Distance from point B = 40 – 21.2293

=18.7707 cm

Resultant force = 203.8633 N ( at an angle of 44.6624o in first quadrant)

Distance of resultant force from hinge B = 18.7707 cm