Dynamics Problem: Not Understanding the Acceleration Term

[ltkach2015]

Homework Statement

[/B]
If particles at B and C of equal mass m are connected by strings to each and to points A and D as shown, all points remaining in the same horizontal plane. If points A and D move with the same acceleration a along parallel paths, solve for the tensile force in each of the strings. Assume that all points retain their initial relative positions.

PLEASE SEE ATTACHMENT

Homework Equations

Kinematics- total acceleration:
A = 0(i) + g(-j) + a(j) + 0(k) = Ax(i) + Ay(j) + Az(k)

Mechanics:
∑F - mA = 0;

Magnitude of acceleration due to gravity:
g =9.81

Labeling:
Tension between A & B: Tab, etc

Coordinate System:

x to the right
y upwards
z out of the page

The Attempt at a Solution

Freebody Diagram on left point mass (EQ1)
∑F1 - mA = 0 = Tab*cosd(45)(-i) + Tbc(i) + Tab*sind(45)(j) - mA

Freebody Diagram on right point mass (EQ2)
∑F2 - mA = 0 = Tdc*cosd(45)(i) + Tbc(-i) + Tdc*sind(45)(j) - mA


SOLVING:

x-componenets of (EQ1) and (EQ2):

Tab = Tdc

Substituting Tab for Tdc into (EQ2) y-component:

m( g(-j) + a(j) ) = Tab*sind(45)(j)

Showing y-component of (EQ1):

m( g(-j) + a(j) ) = Tab*sind(45)(j)RESULTING:
=> Tab = sqrt(2)mAy = Tdc
=> Tbc = mAy

note:

A = 0(i) + g(-j) + a(j) + 0(k) = Ax(i) + Ay(j) + Az(k)
Ay = -g + a

My Questions:
1) The sign of acceleration due to gravity (g) is negative while on the external forces side of the equation, when usually it is negative. Which tells me that there is something wrong with my acceleration term.

2) But, how else can we solve this problem without including the acceleration due to gravity in the total acceleration? Else, I would need to totally neglect it from the solution.

3) If neglected is that the same thing as considering only the dynamic case? i.e. forget the weight = mg. Why? Is there a more intuitive approach?Thank you.

 

[Doc Al]

1) The sign of acceleration due to gravity (g) is negative while on the external forces side of the equation, when usually it is negative. Which tells me that there is something wrong with my acceleration term.

g is a positive constant that describes the strength of the gravitation field.

2) But, how else can we solve this problem without including the acceleration due to gravity in the total acceleration? Else, I would need to totally neglect it from the solution.

The acceleration is given as "a"; g is not part of that.

 

[DEvens]

You are forgetting part of the problem statement, namely the part about "all points remaining in the same horizontal plane." Horizontal, right? So which way does gravity point? So these things must all be taking place on a table of some kind. Right? No gravity involved.

For future reference, you get these things with the sign of things like gravity correct by remembering they are vectors. That is, they have a magnitude and a direction. So gravity points "down." You pick the direction that will be positive, and that tells you the sign.

 

[ltkach2015]

g is a positive constant that describes the strength of the gravitation field. The acceleration is given as "a"; g is not part of that.

Ok I would agree with you too, however, the problem forces me to do one of two things to get the same result:

1) ignore the weight term. I think some call it the dynamic case.
or
2) the weight term is in fact apart of the total acceleration.

Answer:

Tab = Tdc = sqrt(2)*m*a

 

[ltkach2015]

You are forgetting part of the problem statement, namely the part about "all points remaining in the same horizontal plane." Horizontal, right? So which way does gravity point? So these things must all be taking place on a table of some kind. Right? No gravity involved.

For future reference, you get these things with the sign of things like gravity correct by remembering they are vectors. That is, they have a magnitude and a direction. So gravity points "down." You pick the direction that will be positive, and that tells you the sign.

I think you are totally right about it being on a horizontal plane. This take care of the solution. Wow. Thanks.

I don't know what you mean about the directions, I believe I did show them.

 

[Doc Al]

Ok I would agree with you too, however, the problem forces me to do one of two things to get the same result:

1) ignore the weight term. I think some call it the dynamic case.
or
2) the weight term is in fact apart of the total acceleration.

You certainly do not want to ignore the weight of the masses! But that doesn't mean g is part of their acceleration.

Answer:

Tab = Tdc = sqrt(2)*m*a

I'd say that answer is incorrect.

Edit: I misread the problem. Since everything is in a horizontal plane, g plays no role.

 

[Doc Al]

I think you are totally right about it being on a horizontal plane.

D'oh! I missed that. (I had assumed they meant that the masses remain in a horizontal line.)

Good catch, DEvens.