Yesterday, a friend on QQ asked me to help solve a physics problem:
A uniform wooden rod of length L and weight Q has a weight of Q/2 placed on it at a distance of L/4 from end A. When in equilibrium, what is the angle \theta between the wooden rod AB and the horizontal plane?
It looks quite interesting. I noted it down and thought about it for a while this morning, arriving at the following result. In my solution, I did not use direct force analysis; instead, I used the "Principle of Minimum Potential Energy" that we have discussed before: The potential energy in an equilibrium system must take an extremal (minimum) value.
First, as shown in the figure, let us generalize the problem. Let the length of the two thin ropes be b, the length of the wooden rod be L, and its mass be m. Let the mass of the block be M, and its distance from point A be a. Let the coordinates of the center point be (x, y). It is not difficult to establish that x^2 + y^2 = b^2 - (\frac{L}{2})^2 = r^2. Thus, we can set y = r \cos\theta and x = r \sin\theta (this differs slightly from standard polar coordinates, but it is chosen for computational convenience). It is evident that the only variable is \theta, making this a one-degree-of-freedom problem.
Choosing the plane where point O lies as the reference plane, it is easy to see that the potential energy of the wooden rod is E_{p1} = -mgr \cos\theta, while the potential energy of the block is E_{p2} = -Mg[r \cos\theta + (\frac{L}{2} - a) \sin\theta]. The total potential energy is: E_p = -mgr \cos\theta - Mg[r \cos\theta + \left(\frac{L}{2} - a\right) \sin\theta]
Taking the derivative and setting it to zero: \frac{dE_p}{d\theta} = mgr \sin\theta + Mg[r \sin\theta - \left(\frac{L}{2} - a\right) \cos\theta] = 0
Solving for \theta: \tan\theta = \frac{M(L - 2a)}{2r(M + m)} where r = \sqrt{b^2 - \left(\frac{L}{2}\right)^2}.
For the original problem proposed by the QQ friend, we obtain: \tan\theta = \frac{\sqrt{3}}{18}
We can verify the answer from several qualitative perspectives:
Check if the dimensions on both sides are equal. In this answer, both sides are dimensionless, which is clearly satisfied.
When a = \frac{L}{2}, the rod clearly will not tilt to either side; the answer satisfies this.
When a > \frac{L}{2}, the rod will clearly tilt to the other side; the answer satisfies this (\tan\theta becomes negative).
When M is 0, the tilt angle is clearly 0. When M is much larger than m, the tilt angle becomes independent of M and m. The answer satisfies this.
When b is very long (tending toward infinity), the tilt angle will clearly not be significant, so \theta tends toward 0. The answer also satisfies this.
In many cases, verification does not mean recalculating everything from scratch, but rather finding special cases to validate the result as shown above. Because a correct answer must satisfy all aspects of the physical phenomenon. These special cases actually describe the properties of the physical system. This is a qualitative method. Mastering qualitative methods allows us to "grasp" various properties of the answer, or even "guess" the answer, even when we are completely unaware of the specific details! This is precisely what many physics masters like Fermi, Feynman, and Wheeler were skilled at.
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