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If Newton II is defined as ##\sum F = \dot{p}## and ##p = mv##, why do we consider Newton I as a separate law for cases where ##\sum F = 0##? Is Newton I really independent of Newton II?
Newton's first law is spelt out to repudiate the Aristotelian position that objects will naturally come to rest. With that out the way, Newton's 2nd law explains how they actually behave.If Newton II is defined as ##\sum F = \dot{p}## and ##p = mv##, why do we consider Newton I as a separate law for cases where ##\sum F = 0##? Is Newton I really independent of Newton II?
So there is no strict reason we couldn't state that there are only 2 laws of motion instead of 3?Newton's first law is spelt out to repudiate the Aristotelian position that objects will naturally come to rest. With that out the way, Newton's 2nd law explains how they actually behave.
Often the first law is considered a definition of inertial reference frames and the second law is considered a definition of forces.If Newton II is defined as ##\sum F = \dot{p}## and ##p = mv##, why do we consider Newton I as a separate law for cases where ##\sum F = 0##? Is Newton I really independent of Newton II?
That makes sense, but I don't see why we couldn't attribute both of these definitions to the second law.Often the first law is considered a definition of inertial reference frames and the second law is considered a definition of forces.
The second law only holds in a reference frame where the first law holds.That makes sense, but I don't see why we couldn't attribute both of these definitions to the second law.
Hmm, maybe it is possible, but I don’t see an obvious way (and I haven’t seen anyone do something like that). You need to define an inertial frame (so that acceleration is defined) and force.That makes sense, but I don't see why we couldn't attribute both of these definitions to the second law.
I don’t know how without an independent definition of either an inertial frame (needed to define acceleration) or force.I think the second law implies the first, but the converse is not true.