2D Rigid Body Dynamics - Newtonian Equaitons of Motion

[mm391]

Homework Statement

In the diagram attached a bar weighs 500 kg and is a 2D plane rigid body with mass centre at G. At the instant shown, the bar is moving horizontally at but reducing speed, causing horizontal deceleration as shown by vector a=-7i m/s². At the same time, the bar is being raised rapidly by anti-clockwise rotation of the support structure with angular velocity of magnitude ωr=2rad/s. All other kinematic components can be ignored. Assuming the forces at B can only be applied in the normal direction and that the support at point O is a frictionless pin joint construct Newtonian equations of motion, and use just the moment equation about Point O to calculate the instantaneous force in the hydraulic ram at point B.

Homework Equations

∑F_x = m*a_x

∑F_y = m*a_y

∑M_o = I*$\alpha$+m*ad

The Attempt at a Solution

I can see using the equations below that I$\alpha$=0 as there is no angular acceleration. I am struggling to work out what the acceleration component is in the "m*a*d" part of the equation. Can anyone please have a go at explaining (not solving) how I can find it?

Thanks

Mark

 

[Simon Bridge]

The rod is rotating anticlockwise, at a constant angular speed, about one end as that end moves to the right and decelerates.

Presumably there is an unbalanced centripetal force on the rod com that points at O.

Anyway:
Can you state in words what the "mad" part of the moment sum represents?
You may need to consult your notes.

Looks like "a" is the linear acceleration of the center of mass and "d" is the perpendicular-to-a distance from the pivot to the center of mass.

 

[mm391]

Solved. Thanks

 

[Simon Bridge]

Well done :)

 

[rezeew]

First, let's define our coordinate system and variables. Let the x-axis be horizontal and the y-axis be vertical, with the origin at point O. Let the acceleration of the bar be represented by a vector a = -7i m/s^2. Let the angular velocity of the support structure be represented by ωr = 2 rad/s. Let the distance from point O to point G be r.

Now, we can use the given information and the equations of motion to solve for the force at point B. We know that the force at point B is acting in the normal direction, so it can be represented by a vector B = Bn n, where Bn is the magnitude of the force and n is the unit vector in the normal direction.

First, we can use the equation ∑Fy = m*ay to solve for Bn. Since there is no acceleration in the y-direction, the equation becomes Bn = mg. We also know that the acceleration in the x-direction is given by a = -7i m/s^2. We can use this information to solve for the acceleration component in the "m*a*d" part of the moment equation.

Using the equation ∑Fx = m*ax, we can solve for the acceleration component in the x-direction, which is given by ad. This becomes ad = ma = (500 kg)(-7i m/s^2) = -3500i N.

Now, we can use the moment equation ∑Mo = I*\alpha+m*ad to solve for Bn. Since the angular acceleration is zero, the equation becomes ∑Mo = m*ad = Bnr. Plugging in the values we know, we get Bnr = (500 kg)(-3500i N)(r) = -1750ri Nm.

Finally, we can solve for Bn by dividing both sides by r, which gives us Bn = -1750i N. So the force at point B is -1750i N, meaning that it is acting in the negative x-direction and has a magnitude of 1750 N.

I hope this helps in understanding how to approach this problem. Remember to always define your variables and use the equations of motion to solve for the unknowns. Good luck!