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Howers
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Why is it that the two pendulums must have equal masses if complete energy transfer is to take place, assuming the angular frequencies are commnesumerable.
Equal Masses for Complete Energy Transfer in Coupled Pendulums
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Complete energy transfer between coupled pendulums requires equal masses due to the nature of their coupled differential equations. When the masses are equal, the system can be simplified, leading to an uncoupled system that is easier to analyze and integrate. The discussion references classical mechanics principles, suggesting that established texts like Landau's can provide further insights. The assumption of equal masses is crucial for achieving the desired energy transfer dynamics. Understanding these relationships is essential for solving problems involving coupled oscillators effectively.
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Hello!
I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below.
Here is how I wrote the condition of equilibrium forces:
$$
\begin{cases}
F_{g\parallel}=F_{t1}+F_{t2}, \\
F_{g\perp}=F_{r1}+F_{r2}
\end{cases}.
$$
On the other hand...
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