Dynamics - Object Sliding Along A Frictionless Wall

AI Thread Summary
The discussion revolves around determining the angular acceleration and reaction forces at points A and B for a 500lb box released from rest against a frictionless wall. The participant initially used force and moment equations but encountered difficulties with negative reaction forces, particularly at point A. Suggestions were made to utilize conservation of energy to simplify the calculations, focusing on the relationship between linear and angular velocities. Ultimately, it was clarified that a negative reaction force at A implies it can be considered zero, leading to a reevaluation of the calculations. The conversation highlights the importance of understanding kinematic constraints and the relationships between different types of motion in solving dynamics problems.
yoft
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Homework Statement


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The 500lb box is released from rest in the position shown. The wall and ground are smooth. Determine (a) the angular acceleration of the box, and (b) the reactions at corners A and B at the instant after release.

**G = center of mass
**L = length of side = 2.5ft
**phi = 15 degrees (found through geometry)

Homework Equations



ƩF = mag

ƩM = Igα

aa = ab + α x ra/b

The Attempt at a Solution



Okay so the two assumptions I'm working off of here are that points A and B have kinematic constraints, the constraints being that they have to travel along their respective walls; A can only travel downward (-j) and B can only travel sideways (+i) the instant they are released. So these are the two equations I started with:

ƩF = FA(i) + (FB-mg)(j) = maG

ƩMG = (√2/2L)sin(phi)FA+(√2/2L)sin(phi)FB = 1/6mL2α

I then solved for α using the sum of the moments and put it into this equation:

aA = aG + α(k) x rA/G

Here, if A can only accelerate in the j direction, then that means its i terms sum to zero so I pull out the i terms:

{the first term comes from my F=ma equation, aG}
(FB - mg)/m + (6d2cos2(phi)(FA+FB))/(mL2) = 0

I then use the same relative acceleration equation with point B and the same kinematic constraints of all the j terms in this equation summing to zero since B can only move in the i direction:

FA/m + (6d2cos2(phi)(FA+FB))/(mL2) = 0

Now I have two equations and two unknown reaction forces so it should be straight forward, but I get a negative value for FA... I can't help but think that it boils down to a sign error or a faulty cross product somewhere but I've looked over it for 3 hours and I can't find anything mechanically wrong with it. Also, a friend told me that the reaction force at A should be zero, but I'm not sure if he's correct on this. It kind of makes intuitive sense that it would be zero but I can't for the life of me figure out how to mathematically get a positive value for FA, much less get a zero value for it. Any help on this would be immensely appreciated. I'm sorry for the long post.
 
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welcome to pf!

hi yoft! welcome to pf! :smile:

(i haven't checked your equations, but …)

it would be a lot easier to use conservation of energy first, to find the speed and acceleration,

and then to have a go at the reactions! :wink:
 
I actually had tried using energy conservation but I couldn't really get anywhere with it. It's pretty straight forward to use the distance that the center of mass drops to calculate the linear speed right before it hits the floor but I don't know what to do with that information. I suppose the magnitude of acceleration is just gravity but I don't know which direction it's pointing in. I also am unsure as to how I would relate that to alpha. I know a = alpha cross r but isn't r the distance between the instance center and the center of mass? And I don't know how I would find the instance center.
 
hi yoft! :smile:

for conservation of energy, you only ned to know the (linear) velocity of the centre of mass, and the angular velocity

for the angular velocity, just use |vA - vB|/(AB) for any two points A and B

(it works because |vA - vB| = |ω x (rA - rB)| = ωAB)

(in this case, i'd use the two obvious corners!)

try again! :smile:
 
I guess I still don't really understand what you're suggesting :( it seems to me that doing that adds in 2 additional unknowns (linear velocity at points A and B) while not really moving towards getting angular acceleration? And thanks so much for your help and time!

Also: is your "AB" term the magnitude of the distance between points A and B?
 
hi yoft! :smile:
yoft said:
I guess I still don't really understand what you're suggesting :( it seems to me that doing that adds in 2 additional unknowns (linear velocity at points A and B) while not really moving towards getting angular acceleration?

but they're not additional unknowns, they're "instead-of" unknowns! :rolleyes:

the coordinates of A and B completely determine the position of the block, don't they? :wink:

use them to determine everything else! :smile:

(and yes, AB is the distance)
 
Okay so using conservation of energy:

T1+V1+U1-->2 = T2+V2

I get:

mgh = 1/2mvG2 + 1/2IG[(VA-VB)/L]2

I'm guessing I have to put VA and VB in terms of VG somehow and solve for VG? And even if I get VG I'm still not sure how I would relate that to α and the reaction forces. I guess I don't really understand the direction you're showing me :(
 
yoft said:
I'm guessing I have to put VA and VB in terms of VG somehow and solve for VG?

no!

call the coordinates of A and B x and y,

and express everything in terms of x and y​

(and I'm going out for the evening now :wink:)
 
I think I got it, I used the equations of relative accelerations first to get linear acceleration in terms of alpha and then put them into the sum of moments and forces to some for alpha. Thanks for your help though!
 
  • #10
dalen2000 said:
http://www.infoocean.info/avatar1.jpg it would be a lot easier to use conservation of energy first

So I'm told, but I couldn't (and still don't) understand how to go from linear velocity and angular velocity to angular acceleration and force. And I don't know how much easier it really would have been, it took around 6 lines the way I did it.
 
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  • #11
It actually turns out that the reaction force at A is supposed to be negative... And then you're supposed to assume that that means the reaction force at A is actually zero since it can't be negative, and then you redo the calculations under the assumption that A is zero. I don't really know how I feel about being supposed to make that assumption but it's good to know that my answer was right the first time because I couldn't understand why conceptually it was the wrong approach.
 
  • #12
yoft said:
It actually turns out that the reaction force at A is supposed to be negative... And then you're supposed to assume that that means the reaction force at A is actually zero since it can't be negative, and then you redo the calculations under the assumption that A is zero. I don't really know how I feel about being supposed to make that assumption

it isn't assuming that A is zero so much as assuming that the wall isn't there! :biggrin:

… but it's good to know that my answer was right the first time because I couldn't understand why conceptually it was the wrong approach.

yes, either way is correct :smile:

(conservation of energy is usually simpler, but in this case since only the initial forces were asked for, the direct approach was probably quicker)
 
  • #13
Thanks a lot for your time!
 
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