fluid dynamicsfluid-staticsstabilitytorque
I don't understand how does a rotating moment caused by the weight force form when a submerged object is tilted like this. I understand that the buoyancy force produce a moment about CG because of the horizontal distance between FB and CG. But for the weight how can it rotate the object about its center of gravity when the Force is directed through the Center of gravity . How would it create a torque? A free body always create a torque about its CG.
All your thinking is very good and correct. BUT you are missing a point in the question :)
Yes, the door's weight (which pulls from the center og gravity - no need to think about each particle of the door) creates a torque. The hinges then gives a counter torque. True.
BUT let's read the question:
A door is hinged at one end and is free to rotate about a vertical axis. Does its weight cause any torque about this axis?
Does the weight try to turn the door about a vertical axis? Nope! It tries to turn it about a horizontal axis. See the picture below.
To the comment:
There are several forces on the door. All of those cause torques about some axis. But when you choose an axis to look at, a torque will always be zero, no matter the size of the force, in precisely two cases:
It's all about what axis you look at. About a horizontal axis (that does not go through the center of gravity), you are right, there is a torque from the weight, which the hinges have to hold back. This causes stress in the hinges. But about any vertical axis, the weight is unable to make anything turn.
If we take the free body diagram above and sum the moments about the center of mass, we would find that an increased applied force would in fact cause the solid body to rotate.
Perhaps. Or additional forces can appear. If I push up on my car's bumper, a rotational force is being applied. But the normal force on the wheel farther from me increases, so the total torque still sums to zero.
The problem with this understanding is that during rotation, the center of mass of the drawn system should actually rises relative to the surface the bike is moving across. If the bike were to be truly rotating about its center of mass, then the back wheel would begin to dip below the surface of the road like you might see in a glitchy video game.
Another way to interpret this is that if you apply a small torque and imagine it around the center of mass, you're pushing the rear wheel into the ground. As you do so, the normal force increases. This increased normal force counters the torque you are applying. But the maximum this can be is the weight of the bike. So if you increase past this maximum, the bike will rotate.
I had originally tried to analyze this from the rear axle, but because the bike will accelerate, this makes fictitious forces appear in the axle's frame that have to be dealt with. We can mostly ignore this by analyzing around the center of mass instead.
I'll ignore friction for now, and just assume that we have sufficient friction to avoid wheel slip. Then the forces we need to consider are the weight, the normal forces, and the frictional force from the road.
As the wheel accelerates faster it provides a torque to the bike. The bike responds by changing the balance of the normal forces. At the limit, only the rear wheel is providing a normal force, and that will equal the weight of the bike. Since gravity acts through the center of mass, the only torques that appear are the normal force and frictional force. When torque from friction exceeds torque from the normal, the bike will tip.
$$\tau_{friction} > \tau_{Normal}$$ $$F_f \times y > F_N \times x$$ $$F_f > \frac{mgx}{y}$$
To tip the bike, the wheel has to push with a force greater than the weight of the vehicle times a factor that depends on the location of the center of mass. \
And since we have the forward force, we can solve for forward acceleration and know that as it begins to tip, the bike will be accelerating at $\frac{x}{y}g$
And then the bit that I think began your question:
...the only forces acting about the rear axle, point O, is the force of weight. This means that an increase in applied torque and subsequently, the applied force Fa, should not effect the rotation about the rear axle.
In your initial diagram, you were neglecting one additional force, and that is the fictitious force due to acceleration of the frame. This force is equal to $ma$ and is applied to the center of mass in the direction opposite the acceleration of the bike. As the acceleration is due to this force, it means it does affect the rotation, even though it wasn't obvious when you started summing torques.
Best Answer
Actually, the upward buoyant force and downward weight, which are parallel equal but opposite forces, form what is called a "couple". A couple causes pure rotation without translation (since the net force is zero). The magnitude of the couple is the sum of the torque contributions of each of the two forces, or
$$M_{C}=F\frac{d}{2}+F\frac{d}{2}=Fd$$
where $d$ is the perpendicular distance between the lines of action of the two forces, and $F_{B}=W=F$ for a neutrally buoyant body.
When the ensuing rotation caused by the couple vertically aligns the two forces (after some possible oscillations), the body will be rotational equilibrium.
Hope this helps.