Answer to Question #172255 in Mechanical Engineering for waheed

Question

The Lift Mechanism for the package moving is desired to raise (your registration number)                                        

inch in 2.50sec, remain stationery for 2.0 sec, and return in 2.0 sec. The push mechanism 

should remain stationery for 2.50 sec, push (your registration number)inch in 2.0 sec and

return in 2.0 sec. Determine

                 (a) The Time Ratio, Cycle Time                           

                 (b)Sketch the Synchronized time charts

Answer 1

(a) Q(Time ratio) = Time of slower stroke / Time of quicker stroke

Q_{lift mechanism} = 2.5 sec / 2.0 sec = 1.25

Q_{push mechanism} = 2.0 sec / 2.0 sec = 1.00

$\Delta$t_cycle(total cycle time) = Time of slower stroke + Time of quicker stroke

$\Delta$t_cycle(lift mechanism) = 2.5 sec + 2.0 sec + 2.0 sec = 6.5 sec

$\Delta$t_cycle(push mechanism) = 2.5 sec + 2.0 sec + 2.0 sec = 6.5 sec

(b) Motion Parameters (peak velocity and acceleration) of the Lift Mechanism

v_peak = 2$\Delta$R/$\Delta$t₁ = 2*(10.0in)/(2.5 s) = 8.0 in/s

a = 4 $\Delta$R/$\Delta$t₁² = 4*(10.0 in)/(2.5 s)² = 6.4 in/s²

Motion Parameters of the return stroke of the Lift Mechanism

v_peak = 2$\Delta$R/$\Delta$t₁ = 2*(-10.0in)/(2.0 s) = -10.0 in/s

a = 4 $\Delta$R/$\Delta$t₁² = 4*(-10.0 in)/(2.0 s)² = -10.0 in/s²

Motion Parameters of the push mechanism

v_peak = 2$\Delta$R/$\Delta$t₁ = 2*(10.0in)/(2.0 s) = 10.0 in/s

a = 4 $\Delta$R/$\Delta$t₁² = 4*(10.0 in)/(2.0 s)² = 10.0 in/s²

Motion Parameters of the return stroke of the push mechanism

v_peak = 2$\Delta$R/$\Delta$t₁ = 2*(-10.0in)/(2.0 s) = -10.0 in/s

a = 4 $\Delta$R/$\Delta$t₁² = 4*(-10.0 in)/(2.0 s)² = -10.0 in/s²

Time charts: