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Force is directly proportional to mass and velocity and inversely proportional to time so why don't we write $F=1/t+m+v-v_0$ where $m$ is mass, $v$ is final velocity, and $v_0$ is initial velocity?

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    $\begingroup$ How do you add 1/s and kg and m/s?Price of some wood is proportional to length L , thickness d and mass m so why not pay Price P=L+d+m $\endgroup$
    – trula
    Dec 31, 2022 at 18:19
  • $\begingroup$ @trula that's what am asking why we multiply and not just simply add? $\endgroup$ Dec 31, 2022 at 18:27
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    $\begingroup$ You can not add apples and pears in your case any force applied 1s on no mass with nor velocity would be 1 in the unit 1/s? $\endgroup$
    – trula
    Dec 31, 2022 at 18:31
  • $\begingroup$ $1/t+m+v-v_0$ is a dimensional mess. Do you mean $1/t\times m\times(v_f-v_0)=m(v_f-v_0)/t$? $\endgroup$
    – J.G.
    Dec 31, 2022 at 19:39
  • $\begingroup$ the equation in the title is dimensionally incorrect $\endgroup$
    – jim
    Dec 31, 2022 at 20:18

3 Answers 3

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Force is directly proportional to mass and velocity and inversely proportional to time so why don't we write $F=1/t+m+v-v_0$

As others mentioned, the units don’t work. However, suppose we modify it to $$F=k_t/t+k_m m+k_v(v-v_0)$$ where the various $k$ are constants with appropriate dimensions that make each term a force.

Now, you have an equation that is dimensionally consistent. However, it is not directly proportional to mass and velocity and inversely proportional to time. If you double $m$ then according your formula you do not double $F$. Instead you get $$k_t/t+k_m 2m+k_v(v-v_0) \ne 2 F$$

For $F$ to be proportional to $m$ means $F=km$, and similarly with the other factors.

Also, one nitpick. Force is not proportional to velocity but the average force is proportional to the change in velocity. Those are slightly different statements.

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One way to convince yourself of this is that force is measured in $\rm{kg\cdot m/s^2}$ which is a Newton. This can be formed by multiplying mass kg times velocity m/s divided by time s. You cannot get there by adding the units.

More generally, you cannot add any units (or the corresponding dimensioned quantities) unless they are the same. You can add two masses

1 kg + 2 kg = 3 kg

Or two velocities

2 m/s + 4 m/s = 6 m/s

Or two forces

1 $\rm{kg\cdot m/s^2}$ + 0.5 $\rm{kg\cdot m/s^2}$ = 1.5 $\rm{kg\cdot m/s^2}$

But you cannot add a mass and a length

1 kg + 2 meters =.....

It doesn't work.

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By Newton's second law force is defined as mass times acceleration. Acceleration is defined as the time derivative of velocity, and velocity is defined as the time derivative of position. Force, acceleration, velocity, and position are vectors.

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