Taking moments about which axis?

[eurekameh]

Why doesn't taking moments about an axis passing through point A or B work? Why must it be the center of mass here?

 

[andrien]

there are moment of inertia forces when you choose pt A and B.about centre of mass this is zero becuse they pass through it

 

[eurekameh]

Not all forces are passing through the center of mass. Anyone else?
I understand why sum of moments about the center of mass is zero (the refridgerator is not rotating). But why doesn't summing all moments about point A or B work?

 

[PhanthomJay]

You get incorrect results when summing moments about A or B = 0 when the object is accelerating, due to its net force horizontally causing acceleration of its center of mass. You can sum moments about points A or B and get correct results only if you introduce the concept of a ficticious (pseudo) inertial force, P, where P = ma, applied through the object's center of mass horizontally in a direction opposite to the acceleration. This is what Andrien was hinting at. By introducing this concept, the system equates to one that is in static equilibrium, and you can then apply sum of moments = 0 about any point, including the moment from the inertial force. Generally speaking, the use of ficticious inertial forces shoul be avoided, but the concept works well in cases like this. This is known as D'Alembert's Principle (of inertial forces), which I've copied below from the Wiki site:

"D'Alembert showed that one can transform an accelerating rigid body into an equivalent static system by adding the so-called "inertial force" and "inertial torque" or moment. The inertial force must act through the center of mass and the inertial torque can act anywhere. The system can then be analyzed exactly as a static system subjected to this "inertial force and moment" and the external forces. The advantage is that, in the equivalent static system one can take moments about any point (not just the center of mass). This often leads to simpler calculations because any force (in turn) can be eliminated from the moment equations by choosing the appropriate point about which to apply the moment equation (sum of moments = zero). Even in the course of Fundamentals of Dynamics and Kinematics of machines, this principle helps in analyzing the forces that act on a link of a mechanism when it is in motion. In textbooks of engineering dynamics this is sometimes referred to as d'Alembert's principle."

 

[eurekameh]

So in general, when an object is not accelerating, we can sum moments = 0 about any axis, but when the object is accelerating, we sum moments about its center of mass if we do not intend to use inertial forces?
It's weird because my dynamics textbook by Bedford does not introduce the idea of d'Alembert's principle of inertial forces or how we can transform an accelerating system into one that is static.

 

[PhanthomJay]

So in general, when an object is not accelerating, we can sum moments = 0 about any axis, but when the object is accelerating, we sum moments about its center of mass if we do not intend to use inertial forces?

yes

It's weird because my dynamics textbook by Bedford does not introduce the idea of d'Alembert's principle of inertial forces or how we can transform an accelerating system into one that is static.

Neither did mine.

 

[andrien]

in general it is legitimate to take moment about any axis and equate the moment of external forces to the moment of rate of change of momentum.then you need not to know anything about d'alembert's principle.

 

[PhanthomJay]

Yes, but the moment (M) of the change in momentum of the center of mass (m) about any point , O, where the perpendicular distance from the line of action of that momentum change through the COM (i.e, the net force) to the point is y, is
M = (m)(a)(y) , which is precisely the same in magnitude as the moment of the (D"Alembert) inertial force about that point. The only difference is that such moment appears as positive in one case, and negative in the other, because that moment term is on different sides of the equation. The important thing to note in either case is that the inertial force (-F_net = -ma) or the change in momentum force (F_net = ma) , acts through the COM.

 

[deaniscool125]

Hey, this is my first post and I have been forwarded to this page by a friend as I have a similar question to answer.. it can be found here:
http://s1162.photobucket.com/albums/q536/Deaniscool125/
I understand how to take the moments about the centre of mass, however the moment equation then contains both A and B variables. How can I solve for A or B when 2 variables exist?

This may be me being silly but any help will be most appreciated

 

[azizlwl]

You get incorrect results when summing moments about A or B = 0 when the object is accelerating, due to its net force horizontally causing acceleration of its center of mass. You can sum moments about points A or B and get correct results only if you introduce the concept of a ficticious (pseudo) inertial force, P, where P = ma, applied through the object's center of mass horizontally in a direction opposite to the acceleration. This is what Andrien was hinting at. By introducing this concept, the system equates to one that is in static equilibrium, and you can then apply sum of moments = 0 about any point, including the moment from the inertial force. Generally speaking, the use of ficticious inertial forces shoul be avoided, but the concept works well in cases like this. This is known as D'Alembert's Principle (of inertial forces), which I've copied below from the Wiki site:

"D'Alembert showed that one can transform an accelerating rigid body into an equivalent static system by adding the so-called "inertial force" and "inertial torque" or moment. The inertial force must act through the center of mass and the inertial torque can act anywhere. The system can then be analyzed exactly as a static system subjected to this "inertial force and moment" and the external forces. The advantage is that, in the equivalent static system one can take moments about any point (not just the center of mass). This often leads to simpler calculations because any force (in turn) can be eliminated from the moment equations by choosing the appropriate point about which to apply the moment equation (sum of moments = zero). Even in the course of Fundamentals of Dynamics and Kinematics of machines, this principle helps in analyzing the forces that act on a link of a mechanism when it is in motion. In textbooks of engineering dynamics this is sometimes referred to as d'Alembert's principle."

New and very good information for me.
I guess it is not wrong to consider the ficticious (pseudo) force including centrifugal force.

 

[PhanthomJay]

Hey, this is my first post and I have been forwarded to this page by a friend as I have a similar question to answer.. it can be found here:
http://s1162.photobucket.com/albums/q536/Deaniscool125/"'
I understand how to take the moments about the centre of mass, however the moment equation then contains both A and B variables. How can I solve for A or B when 2 variables exist?

This may be me being silly but any help will be most appreciated

For the forces acting on the car of color when accelerating at 1.5 m/s^2 before the collision, first determine the net force acting on it in the horizontal direction using Newton 2. Now apply a pseudo force at the center of mass which has the same magnitude as the net force but which points in the opposite direction. Identify and show the real forces acting on the vehicle in the vert and horiz directions, including its weight, normal, and friction forces , both known and unknown, and you may now sum moments about point A using Newton 1.
NOTE: the friction force acts at the real wheels and is not a function of the static friction coefficient. This problem statement has a lot of holes in it. Also, the use of pseudo forces should in general be avoided like the plague, except in cases like this one.