Question

The linear actuator is designed for rapid horizontal velocity v of jaw C for a slow change in the distance between A and B. If the hydraulic cylinder decreases this distance at the rate u = 28 mm/s, determine the horizontal velocity of jaw C when the angle θ = 49°. The length L = 540 mm.

Answer 1

Answer:

The velocity of jaw is 149.94 mm/s

Explanation:

Here α = θ

The total length in x-axis is:

$x=7(\frac{L}{2}sin\frac{\alpha }{2} )=\frac{7}{2} Lsin\frac{\alpha }{2}$

$v=\frac{7}{2} L*\frac{\beta }{2} cos\frac{\alpha }{2}$

The total length in y-axis is:

$y=Lcos\frac{\alpha }{2}$

The velocity of jaw is:

$v=\frac{7}{2} (\frac{2u}{sin\frac{\alpha }{2} *2} )cos\frac{\alpha }{2} =\frac{7}{2} ucot\frac{\alpha }{2} =\frac{7}{2} *28*cot\frac{49}{2} =149.94 mm/s$

Answer 2

Answer:

99mm/s

Explanation:

Please see attachment for step by step guide