The velocity needs to be determined before the acceleration can be calculated.

The motion of the lift can be determined using the following conditions;

1. Point B only moves vertically
2. Vertical speed of point C is twice that of point B
3. Point C and the lift have the same vertical speed

Velocity Diagram for Member AC

The geometric constraints of points A and B are critical in determining the member AC angular velocity. The velocity vectors for points A and B are

     v_A = v_Ai

     v_B = v_Bj

The relative velocity equation for bar AB is

     v_B = v_A + v_B/A

     v_B = v_A + ω_AB × r_AB

     v_Bj = v_Ai + ω_ABk × r_AB (cosθi + sinθj)

          = v_Ai + ω_AB r_AB (cosθj - sinθi)

Sum in both the i and j directions gives,

     j :    v_B = ω_ABr_ABcosθ

     i :    v_A = ω_ABr_ABsinθ

There are only two unknowns, v_A and ω_AB, which can be solved for, giving

The vertical velocities of points C and B are related by a factor of 2, giving,

     v_C = 2v_B = 1.33 ft/s

Acceleration of Member AC

As with the velocity, the geometric constraints of points A and B are used to determine their acceleration direction,

     a_A = a_Ai

     a_B = a_Bj

The relative acceleration equation for bar AB is

     a_B = a_A + a_B/A

         = a_A + (a_B/A)_n + (a_B/A)_t

         = a_A + ω_AB × (ω_AB × r_AB) + (α_AB × r_AB)

     a_Bj = a_Ai + ω_ABk × (ω_AB k × r_AB (cosθi + sinθj))
               + α_ABk × r_AB (cosθi + sinθj)

     a_Bj = a_Ai + ω²_AB r_AB (-cosθi - sinθj))
              + α_AB r_AB (cosθj - sinθi)

Summing in both the i and j directions gives,

     i :    0 = a_A - α_ABr_ABsinθ - ω²_ABr_ABcosθ

     j :    a_B = α_AB r_AB cosθ - ω²_ABr_ABsinθ

Motion of Platform

These two equations can be used to solve for the two unknowns, α_AB and a_B. Solving for α_AB in the first equation and then substituting into the second equation gives,

Using numerical values give,

     α_AB = 0.04938 rad/s²

     a_B = -0.06959 ft/s²

The vertical acceleration of the lift is twice the vertical acceleration of point B,

     a_lift = 2a_B = -0.1392 ft/s²