F116 The scissors linkage is subjected to a force of P 150

F11-6· The scissors linkage is subjected to a force of P = 150 N. Determine the angle for equilibrium. The spring is unstretched at = 0\". Neglect the mass of the links. sn trethedate he anedertohr cquitbriam The 0.3 m y P= 150 N 0.3 m 11-6

Solution

Solving for static equilibrium
Horizontal force on the vertical bar = P = 150 N
But if Tension in the two members attached to the rod is T
P = T*02sin(theta) = 150
Tsin(theta) = 75 -- (1)

Now, considering the force in the spring
let the compression of the spring be x
then , Tcos(theta) = kx
but, at theta = 0, x = 0
so, x = 0.3 - 0.3*cos(theta) = 0.3(1- cos(theta))
so, from 1 and 2
tah(theta) = 75/k(0.3(1- cos(theta))) = 75/15*0.3(1-cos(theta))
tan(theta) - sin(theta) = 16.6667
sin(theta) - coa(theta)(sin(theta)) = 16.667 cos((theta))
sin(theta) = 16.6667cos(theta)/(1 - cos(theta))
1 - cos^2(theta) = 277.77cos(^2(theta))/(1 + cos^2(theta) - 2cos(theta))
let cos(theta) = y
[1 - y^2][1 + y^2 - 2y] = 277.77y^2
1 + y^2 - 2y - y^2 - y^4 + 2y^3 = 277.77y^2
y^4 - 2y^3 + 277.77y^2 +2y - 1 = 0
after solving this biquadratic equaito the value for y = cos(theta) can be found