MadSci Network: Physics |
Hi Roman, It is always better to try to "prove something is wrong" than to "prove someone wrong". This is because the former point of view allows you to focus on IDEAS and not people. In order to solve the problem I'll reformulate it more formally and I will also make some assumptions not mentioned by you. * all happens in an uniform gravitational field characterized by g constant acceleration towards "DOWN" * the "T" body is constrained so that it can only rotate around an axis that passes thru the 'join' of its 'arms' (horizonatal and vertical); this axis is perpendicular to both arms * this motion is accompanied by 'some' friction * there are two masses Mleft and Mright attached to the heads of the horizontal arm; it does not matter how they are attached ? specify the angle of inclination which corespond to the equilibrium The fundamental law of mechanics (Newton) is: F=ma "Translating" this to rotational motion (such as the motion our T body is confined to): M = J (d/dt)K where M = Force momentum J = Rotational inertia (d/dt)K = variation (in time) of kinetic moment At equilibrium the body does not move so it has zero kinetic moment. And this kinetic moment is constant: it does not vary in time. So the above equation becomes: M = 0 (1) This is the equation used to solve your problem! Only one more thing... How exactly is the force momentum calculated? This is defined in the process of "translating" mentioned earlier. The result is M = F d (2) where F = force magnitude d = distance from the force to the reference point (the 'join' in our case) There are two cases: 1) The T body has no mass In this case if Mleft > Mright the T body will end up in this position: | |----- | Because this is the only position where the momentum of the two weights compensate each other: they are zero because d in equation (2) is zero for both of them. 2) The T body has a mass Mt In order to solve this we must use another important fact from mechanics. The mass of a rigid body can be considered to lie exclusively in a point named center of mass (which is not necesarily 'inside' the body). Now I need to make another hypotesis: * the center of mass of the T body is on the vertical arm, at a distance Lcm from the 'join' * the length of the horizontal arm is 2xLa * the angle made by the 'vertical' arm with the vertical is alpha * Mleft > Mright Now I write the force momentum for the three weights involved here: M1 = + Mleft g La cos(alpha) (3.1) M2 = - Mright g La cos(alpha) (3.2) M3 = - Mt g Lcm sin(alpha) (3.3) The signs are different because the forces tend to rotate the body in different directions. By using equations (3) and (1) you can find that: Mleft - Mright La tan(alpha) = ---------------- ----- (4) Mt Lcm In the first case the angle of inclination does not depend on the masses but in the second case it depends accordingly to (4). PS: Equation (4) pretty much describes how a balance work. Now, I must tell that I'm not sure this is what you've asked. If not then please be more specific. email: radugrigore@ieee.org
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