Simple Statics Problem--what am I missing?

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The discussion centers on the application of torque calculations in a scenario involving a car's acceleration. It emphasizes that while the center of mass (COM) can be used as the axis of rotation for torque, any fixed point in an inertial frame is also valid, provided that pseudo forces are accounted for when necessary. The key point is that Newton’s 2nd law applies to the acceleration of the system's center of mass, and even without rotational acceleration, torques must be summed about the COM. The use of pseudo forces can simplify the analysis by creating an equilibrium situation where all forces and pseudo forces sum to zero. Ultimately, the choice of the COM is justified due to the nature of linear acceleration and the absence of a moment from the pseudo force about the COM.
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Homework Statement
Calculate forces and torques
Relevant Equations
Force and torque equilibrium.
I get the correct answer if I use the COM as the axis of rotation for torque calculations. Shouldn't I be able to use any point as the axis of rotation for torque calculations, e.g. where the front wheels make contact with the ground?
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Newton’s 2nd law applies to acceleration of the center of mass of the system. Even though there is no rotational acceleration, you must sum torques about the COM.
There is a way around this by using the concept of inertial forces, called pseudo forces. Since F=ma, rewrite it as F - ma = 0, where the quantity ma is the pseudo force. So apply the pseudo force ma at the COM , in the direction opposite the acceleration. Now you have an equilibrium situation since all forces and pseudo forces sum to zero, and you can thus sum torques about any point equal to 0 to get the answer.
I don’t often recommend the use of pseudo forces, although it works well here.
 
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Taulant Sholla said:
Shouldn't I be able to use any point as the axis of rotation for torque calculations,
It is not a statics problem: the car is accelerating. You can use the COM or any fixed point in an inertial frame; otherwise you must include pseudo forces as @PhanthomJay describes.
 
PhanthomJay said:
Newton’s 2nd law applies to acceleration of the center of mass of the system. Even though there is no rotational acceleration, you must sum torques about the COM.
There is a way around this by using the concept of inertial forces, called pseudo forces. Since F=ma, rewrite it as F - ma = 0, where the quantity ma is the pseudo force. So apply the pseudo force ma at the COM , in the direction opposite the acceleration. Now you have an equilibrium situation since all forces and pseudo forces sum to zero, and you can thus sum torques about any point equal to 0 to get the answer.
I don’t often recommend the use of pseudo forces, although it works well here.
This is very helpful, thank you so much!
 
haruspex said:
It is not a statics problem: the car is accelerating. You can use the COM or any fixed point in an inertial frame; otherwise you must include pseudo forces as @PhanthomJay describes.
Yes, that does clear things up--thank you!
 
haruspex said:
You can use the COM or any fixed point in an inertial frame;
why is that so? What makes valid the choice of COM for torque calculations in this problem?

I know that if two points positions vectors differ by ##\vec{d}## then the torques around those points differ by ##\vec{d}\times \sum \vec{F}## so if the sum of forces is not zero, then the total torque depends on the point chosen. I just don't understand why choosing COM (or any fixed point in an inertial frame) works for problems like this.
 
Delta2 said:
What makes valid the choice of COM for torque calculations in this problem?
Because the problem is the linear acceleration of the body, and the pseudoforce representing that has no moment about the CoM.
 
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