The discussion revolves around the expression for force, specifically the equation $$F=\frac{mdv}{dt}=\frac{dmv}{dt}$$. Participants explore the mathematical justification for manipulating this equation, considering both constant and variable mass scenarios. The scope includes mathematical reasoning and conceptual clarification regarding the application of the equation in different contexts.
Participants generally agree on the validity of the equation when mass is constant, but there is disagreement regarding its applicability in cases where mass is not constant, such as in rocket dynamics.
The discussion highlights the dependence on the assumption of constant mass and the implications of variable mass in certain physical scenarios, such as rocket motion. Unresolved mathematical steps related to the differentiation of mass are also noted.
You can either simply say because ##m## is a constant, i.e. ##m## doesn't vary in time, or you could do it the long way with the Leibniz rule ##\frac{d}{dt}(m\cdot v)= (\frac{d}{dt}m)\cdot v + m \cdot \frac{d}{dt}v = 0 \cdot v + \frac{m \cdot dv}{dt}## which also uses that ##m## is independent of time, such that its differentiation along time becomes zero. So whether one uses ##\frac{d}{dt}(c \cdot F(t))= c \cdot \frac{d}{dt}F(f)## or ##\frac{d}{dt}c = 0## doesn't matter. As long as the mass doesn't vary in time, the two expressions are equal.OcaliptusP said:The question i want to ask is why
$$F=\frac{mdv}{dt}=\frac{dmv}{dt}$$
I mean how can we move we in the fraction?
Is there are mathematical proof or comes from something else?