The weight of the added water is equal to what the bouyance force was, and the line of action is where the center of the immersed part of the rod used to be.
I am really qlueless as for the line of action of the force. I know the magnitude of the bouyancy force it is equal to the weight of the displaced water which is the half of the volume of the rod and times 1000 kg/m^3. The line of action is important because this a torque problem and since the...
Homework Statement
Lets consider a rod of length l that is attached at a point to the cealing and relvolves freely. Half of the rod is immersed in water and half of it is out of the water. The rod is not vertical it has around a 60 degree angle to the water. Where is the line of action of the...
If I take a rod of length L and lay it down on a surface with no friction and act upon it with a force F off its center of mass by let's say L/4, there will be a torque on the object and it will start to rotate around its center off mass. My question is why does the object necessarily rotate...
I have thought some about what you said. You said that the point of contact travels dintence d and the center of mass does not hance the time it was accelerated is less. But that implies that the contact pont always has larger acceleration than center of mass. How can you be sure about that?
And would the difference between the distences done by the center of mass and the point of contact be calculatable with concrete data e.g the rod is 1m its mass is 1 kg the distence is 1m and the force is 1N.
You did when you said that the linear motion of the two cases cannot be the same. Since if the acceleration is F/m in both cases then the distence d is done in the same time in both cases and if the acceleration and the time is the same the linear terminal velocity has to be equal.
But if both accelerate the same linearly than both has the same terminal velocity so they both have the same kinetic energy, but one rotates so it has additional rotational kinetic energy, but we have done the same work on both objects(F*d) so they should have the same energy. Otherwise its...
Yes but if the linear accelerations are different than that contradicts the fact that regardless of the line of action being at or off the center off mass the linear acceleration has to be F/m.
Im sorry for nagging you, but I am clueless regarding what happens to the applied force. I have been thinking about this for weeks and I can not resolve the problem. I would be very greatful if you did so please explain in full detail without giving clues to me beacuse I have falied in solving...
A yes no answer does not really help me at all. Could you please explain what exactly is wrong with thinking about energy in this though experiment and then why the thought experiment makes sense when considering momentum. Thank you.
Homework Statement
Lets take a long rod of length l and mass m in a zero g vacuum. If a force F acts upon the center of mass of the rod then the rods linear accelaration will be F/m. If the force acts on a distence of d then the work done by the force is F*d. If the force acts off of the...
Homework Statement
Can anyone please give me hard and interesting problems regarding fluid mechanics problems such as hydrostatic and air prssure bouyance force and Bernoullis equation(Bernoullis equation problems only for liquids)? I tried looking for them but the ones I found were quite...