Scissors Lift Equations??
I was looking for some lift equations for our Vex students to use. I derived some myself but decided to check them against your site. The equations stated in this [URL="https://www.engineersedge.com/mechanics_machines/scissor-lift.htm"]engineersedge scissor-lift link[/URL] seem incorrect.
For the bottom actuator the force should be F =( W + wf/2)/tan(phi)
and
for the center pin actuator the force should be F = 2*(W+wf/2)/tan(phi)
Where W is the load weight, wf = the total frame weight and phi is the interior angle between the horizontal and the arm.
The proof used for the bottom force incorrectly assumes that the P force which is the vector sum of F and (W + wf)/2 vertical force is colinear with the arm. This cannot be since there is a moment about the center pin of lift arm that is caused by the force of the load. This requires a component of force at the bottom that is normal to the arm at the point where F is applied. So P cannot be colinear with the arm. If the moment equations are written about the center pin of the lift we get:
F*L*sin(phi) = (W + wf/2)*L*cos(phi)
or F =( W + wf/2)/tan(phi)
This is easily checked by an energy approach. If the load W is lifted by dh then the center of mass of the frame (with weight wf) is lifted by dh/2.
We know that if the actuator moves a distance of dx the work input is F*dx which must equal the potential energy increase of the lift masses moving against gravity.
So F*dx =(W + wf/2)*dh
or
F = (W+ wf/2)*dh/dx
From geometry, dh/dx = 1/tan(phi)
Hence F =( W + wf/2)/tan(phi);
When the force F is applied to the center pin of the lift the same dh is achieved with dx/2 movement so the force F must be twice as large to raise the masses.
So F_center_pin = 2*F_bottom = 2*( W + wf/2)/tan(phi);
Hope you structural engineers set one of us straight.
chris