Why don't we take force proportional to velocity? [duplicate]

When a Force is applied on a body at rest, it produces Acceleration which is equal to change in Velocity. But due to the Force, we see that the Velocity is increasing consequently. So why don't we take both of them proportional to each other rather than Force and Acceleration?

marked as duplicate by tpg2114♦, Brandon Enright, user10851, ACuriousMind♦, Qmechanic♦Dec 15 '14 at 0:19

• for simpler use we use F = ma but 2nd law is F ~ dp/dt, p=mv p- momentum , ~ - proportional to. – Gowtham Dec 14 '14 at 16:14
• Who told you that the force is proportional with the velocity? They don't have to be proportional. The relation between them may be quite complicated sometimes. – Sofia Dec 14 '14 at 16:36
• You may be interested to know that what you are proposing is closely related to to the laws of motion which Aristotle developed. Also, let me ask you this: throw an object into the air; its velocity starts as positive, goes to zero, and then becomes negative, but did the force on it (i.e. its weight) ever change? – Geoffrey Dec 14 '14 at 17:02
• Hey Geoffrey your answer is perfect for me. – user66452 Dec 14 '14 at 18:07
• Posssible duplicates: physics.stackexchange.com/q/87207/2451 and links therein. – Qmechanic Dec 14 '14 at 18:56

You mix the relations between the things. A force produces changes in the linear momentum. The acceleration $a = F/m$ produces changes in velocity $v = p/m$.

So, your question should be either why don't we take the force proportional to the linear momentum, or why don't we take the acceleration proportional to the velocity.

Now, the second thing is proportionality. Sometimes the force is indeed proportional to the linear momentum (the force of friction), and therefore the acceleration that it imposes is proportional to the velocity, and sometimes the relation between force and linear momentum, (analogously between acceleration and velocity), has more complicated expressions.

Well, technically we do: $$F_{net}=m\frac{dv}{dt}$$ The net force is proportional to the rate of change of velocity, which we call acceleration. Note also that it's not $F\propto \Delta v$ (force proportional to the changed velocity because the changed velocity occurs over a period of time, $\Delta t$, that is also important--consider the difference in forces for a 2 m/s change in 1 second versus a 2 m/s change in 1 hour!). We often simplify this to $F_{net}=ma$ because some people are not comfortable with differential calculus.

And also in the case of drag, we say that the force on the falling body that slows it down is, $$F\sim A\rho v^2$$ where $A$ is the surface area of the object, $\rho$ the density of air, and $v$ the velocity of the object. This is how the terminal velocity of an object is determined.

Force was defined to be proportional to acceleration because that definition makes description of classical physics simple. For example on Earth we have a downward pointing constant force - gravity.

If we define another quantity, one which is proportional to velocity, lets call it push, it would be quite useless in describing gravitational interactions. An experiment of dropping a stone from the height of one meter would show that the push of this stone was zero at the starting point and downward pointing in the rest of the trajectory. But if we throw a stone up the push would be upward pointing. Therefore we can't use push to describe how things behave under gravity.

I guess, this quantity could approximate the behavior of a very light object (e.g. a speck of dust) in gas or liquid. Its push would be equal to velocity of the medium at the point with proportionality constant $1$.

If you mean force proportional to velocity, that restricts the second law to only the specific case when the force is proportional to the velocity (the object will feel a drag or will accelerate exponentialy with time, depending on the sign of the proportionality constant). Such a law will not decribe the dynamics of any other objects.