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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.

2 Answers

6 votes

Answer:

vB = 0.5418 m/s (→)

aB = - (0.3189/L) m/s²

ωcd = (0.2117/L) rad/s

Step-by-step 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

The hydraulic cylinder imparts a constant upward velocity vA = 0.23 m/s to corner-example-1
User Azulflame
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7.5k points
1 vote

Final answer:

The question asks for calculations related to the dynamics of a hydraulic cylinder motion mechanism, specifically the velocity and acceleration of a roller and the angular velocity of an edge. However, the provided information does not directly answer the problem but involves similar concepts in rotational motion.

Step-by-step explanation:

The problem involves calculating the velocity vB, acceleration aB of roller B, and the angular velocity ω of edge CD for a hydraulic cylinder mechanism based on given conditions. Unfortunately, the provided information does not relate to the given problem directly but hints at similar concepts that could be used to solve a problem involving related dynamics. In the case of the angular velocity of a cylinder in motion, the equations of rotational motion apply, and kinematic relationships between linear and angular velocities are relevant.

For a cylindrical object, the linear velocity v at the perimeter can be related to its angular velocity ω using the formula v = rω, where r is the radius of the cylinder. The angular acceleration α is connected to the change in angular velocity Δω over time Δt with the formula α = Δω/Δt. To solve the student's initial problem, we would use similar principles, creating a system of equations that relate linear motion at one corner of the container to rotary motion at another point.

User DxAlpha
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8.2k points
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