Rigid Body Kinetics: Find Reaction Force at A

AI Thread Summary
The discussion focuses on calculating the reaction force at point A of a beam system involving a block and a bar. The user attempts to apply fundamental equations of motion and moments but struggles with incorporating the velocity of the block into their calculations. Key insights suggest that the upward acceleration of the bar's end relates to the tension in the cord and the downward acceleration of the mass. The relationship between the angular acceleration of the bar and the upward acceleration at the end is emphasized as crucial for solving the problem. Overall, guidance is provided on how to effectively utilize the tension and acceleration in the moment balance to find the desired reaction force.
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Homework Statement


As the picture says, we are trying to find the reaction force on the bearing at A:
We are given:

M(block) = 26.1kg
M(bar) = 8.7kg
L(bar) = 1.8m
V(block) = 6.2m/s

Homework Equations



What I have attempted the solution with are:

ƩF = ma
ƩM = Iα
I0 = (m*L^2)/12 + m*d^2

The Attempt at a Solution



So, my first approach was to draw a FBD of the mass on the left:

∴ ƩF = ma = T - mg

leading to:

T = m(a+g)

But quickly found out this was useless, so I continued to analyze the bar:

I drew a FBD of the bar with the tension on the very left end, it's gravitational force in the middle and the x and y components of reactions on the far right

I then took the moments around point A to give:

ƩMa (clockwise positive) = Iα = (Tx1.8)-(8.7*9.81*0.9)

Now I calculated I for the beam:

I = (8.7/12)*(0.9^2) + 8.7*(0.9^2)
∴ I = 7.63425

∴ 7.63425*α = (T*1.8)-(8.7*9.81*0.9)

And this is where I am stuck. I know I have to utilize the velocity but I have no idea where that fits into the question, so I went ahead and calculated the angular velocity of the tip of the beam:

ω = v/r = 6.2/1.8 = 31/9

But like I said I have no idea where this fits in.

Any guidence would be super appreciated.

<3
 

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The v appears to be associated with the mass.
This suggests that the beam is pulling the mass upwards.
 
Simon Bridge said:
The v appears to be associated with the mass.
This suggests that the beam is pulling the mass upwards.

Thanks for the reply,

I understand that the beam is pulling the block up, I just don't understand where it fits into the calculation, as I have calculated the corrosponding angular velocity of the bar.

Is there a formula I am missing?
 
The pivot end of the beam is labelled A, so let the other end be B.
If the mass rises with speed v, then how fast does B fall?
What is the relationship between the speed of B and the angular velocity?
 
Let T be the tension on the cord. Do a free body diagram on the mass. In terms of T, what is the acceleration on the mass? The mass is accelerating downward. This must be the rate of acceleration of the end of the bar upward. So you know the upward force at the end of the bar (T), and you know the upward acceleration of the end of the bar. Now do a moment balance on the bar. The angular acceleration of the bar is equal to the upward acceleration of the end of the bar divided by its length. This should give you enough information to solve for T.

Chet
 
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