The hydraulic cylinder imparts a constant upward velocity vA = 0.23 m/s to corner A of the rectangular container during an interval of its motion. For the instant when θ = 23°, determine the velocity vB and acceleration aB of roller B. Also, determine the corresponding angular velocity ω of edge CD. The velocity and acceleration of B are positive if to the right, negative if to the left. The angular velocity of CD is positive if counterclockwise, negative if clockwise.

vB = 0.5418 m/s (→)

aB = - (0.3189/L)  m/s²

ωcd = (0.2117/L)  rad/s

Explanation:

a) Given:

vA = 0.23 m/s (↑) (constant value)

If

tan θ = vA/vB

For the instant when θ = 23° we have

vB = vA/ tan θ

⇒ vB = 0.23 m/s/tan 23°

⇒ vB = 0.5418 m/s (→)

b) If tan θ = vA/vB   ⇒   vA = vB*tan θ

⇒  d(vA)/dt = d(vB*tan θ)/dt

⇒  0 = tan θ*d(vB)/dt + vB*Sec²θ*dθ/dt

Knowing that  

aB = d(vB)/dt

ωcd = dθ/dt

we have

⇒  0 = tan θ*aB + vB*Sec²θ*ωcd

ωcd = - Sin (2θ)*aB/(2*vB)

If

v = ωcd*L

where v = vA*Cos θ   ⇒  ωcd = v/L = vA*Cos θ/L

⇒ vA*Cos θ/L = - Sin (2θ)*aB/(2*vB)

⇒ aB = - vA*vB/((Sin θ)*L)

We plug the known values into the equation

aB = - (0.23 m/s)*(0.5418 m/s)/(L*Sin 23°)

⇒ aB = - (0.3189/L)  m/s²

Finally we obtain the angular velocity of CD as follows

ωcd = vA*Cos θ/L

⇒ ωcd = 0.23 m/s*Cos 23°/L

⇒ ωcd = (0.2117/L)  rad/s

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y