So, I have recently learned about 'Centers of Mass' and have discovered I was completely ignorant to the fact that the point where a force is applied will determine whether the force is a rotational force or a normal force.

Naturally, I have a few questions.

Q1: I have two objects on the ground, that are say at-least 2 cm apart, and center of mass is between the two objects. Does that mean if I applied a net force at the centre of mass the system of objects would move in that direction?

Q2: I have attached an image, in the image, we are looking at 3 objects with masses $5$$kg$, $2$$kg$ and $3$$kg$ in red dots respectively all sitting on a horizontal axis. We are defining $\bar{x}$ as the location of the center of mass from the point A and the total weight of this system is $W=10g N$

Now, in the next line, finding the torque about A and defining clockwise as positive.

$$10g\bar{x} = 5g(2) + 2g(3) + 3g(6) $$

Now, I don't understand how this is so because, I thought that force applied at the centre of mass would move the object(s) in the direction in which the force was applied. So, basically there would be no torque and the weight of the system of objects would just move the system downwards, since weight acts downwards. So $10g\bar{x} = 0$

Image here


Yes, for (1), you're correct.

However, that's assuming that the two objects are joined by some kind of light rod. In that case, applying a force at the center of mass would cause some translational motion.

Regarding (2), your calculation for the center of mass is that the center of mass is 3.4m from A.

Using the formula for torque, $$τ = r * f$$ The 3kg mass is 2.6m from the center of mass, the 2kg mass is 0.4m away from the center of mass and the 5kg mass is 1.4m away from the center of mass.


$$τ_{net} = 3g * 2.6 - 2g * 0.4 - 5g * 1.4 = 0$$

In your calculation, you were taking moments around the point A. What your calculations actually mean is that if all three masses were joined by a rod that was fixed to point A (hence the entire system can only rotate around A), there would be a net torque, which is true.


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