Question

2) The hydraulic cylinder extends and causes point B to have the velocity and acceleration shown. Set up the equations needed to determine the acceleration of point A at the instant shown. Complete all cross products and separate the equations into components. You do not need to solve the equations for the unknowns. CO B us = 4 ft/s ab = 1.5 ft/s

Answer 1

Distance between point B and instantaneous centre for rod AB is given as follows- Polic = AB cos 30º = 2 cos 30º = 1.732 ft Angular velocity of rod AB is given as follows- 04B = _VB - 4 "BLIC 1.737 = 2.309 rad/s Position vector directed from B to A is given as follows- rur = AB cos 30°i – AB sin 30° j = 2 cos 30°i – 2sin 30°i rab = (1.732i – 1j) ft Now acceleration of point A is given as follows- a = ag + Q AB XFAB -(@AB) ralb -a,i =1.5j +(-a ask)(1.732i –1j) - (2.309)*(1.732i –1j)

(1) -a,i=1.5j-1.7320 AB] – ABI – 9.234i +5.331j -a,i=(-0 AB -9.234)i +(6.831-1.7320 AB)j Now equate the components of j on the both side of equation (1). 0 = 6.831–1.7320 AB A AB = 3.944 rad/s Now equate the components of i on the both side of equation (1). -a, =-& AB -9.234 aA = A AB+9.234 aA = 3.944 +9.2343 a = 13.178 ft/s +