Question about Inertia: Force to Overcome Mass of 1kg?

  • Thread starter Thread starter Macrobe
  • Start date Start date
  • Tags Tags
    Inertia
AI Thread Summary
To overcome the inertia of a 1 kg mass, the force required depends on the desired acceleration, as defined by Newton's second law (F = ma). Specifically, 1 Newton of force will accelerate a 1 kg object at 1 m/s². However, if the object is stationary on a flat surface, the force needed to initiate movement must also account for static friction, which is calculated as the coefficient of static friction multiplied by the object's weight. On a typical road surface, this force is approximately 0.7 times the weight of the object. Understanding these dynamics clarifies the relationship between mass, force, and motion.
Macrobe
Messages
12
Reaction score
0
I have a question regarding Inertia. Exactly how much force, in Newtons, is required to overcome the inertia of exactly one kilogram of mass? It seems to me there must be some sort of quantity for it.
 
Physics news on Phys.org
It depends on how fast you want to accelerate the object. The equation is ##\vec{F_{Net}} = m\vec{a}##, or in words, (net force acting on the object) = (mass of the object) x (acceleration of the object). This is Newton's second law. It takes, for example, 1 Newton of force to accelerate a 1kg object at 1m/s2.
 
Ah. Okay...not exactly what I was looking for, but what I was looking for was probably wrong ;) thanks.
 
Yes, if you were looking for some measure of a force required to overcome inertia that is independent of acceleration, then there's no such thing.
 
yeah, it is pretty counter-intuitive that f=ma, because in our everyday lives, most of the time there are frictional forces.
 
I think Macrobe meant to ask how much force would be required to get a body off its stationary position. On a flat surface, it is equal to the co-efficient of static friction x Weight of the body (weight, not mass). If the body is on, say, a road, the force required to get the body would be around 0.7 times the weight of the body.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top