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Imagine a toy car moving with a constant velocity. By Newton's second law of motion we can see that $F=ma$ and, since there is no acceleration, the net force is zero. But still we can experience a force on us downwards. Can anyone explain the logical reason behind this?

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closed as off-topic by Aaron Stevens, Thomas Fritsch, Kyle Kanos, Jon Custer, tpg2114 May 30 at 13:19

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    $\begingroup$ I don't see any attempt to research this. It is a typical homework concept question looking at the vector nature of forces. $\endgroup$ – Bill N May 29 at 11:57
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net force is zero

Not the force is 0. Force in the downward direction is balanced by normal action of floor/road

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This might seem a bit counter intuitive, but we actually don't really experience the force on us downwards. When it is only gravity acting to pull you down (i.e. freefall) without any other forces, you cannot feel that gravity is acting on you. It feels the same as floating without any forces.

What the toy car actually experiences is the upwards force from the ground pushing back on it. So it's somewhat non-intuitive, but you could say in this case the logical reason that you feel the ground pushing you is because you don't feel the force of gravity on it's own. You only feel the reactions to the gravitational force. In this case, the reaction is the normal force from the ground, which prevents the car from accelerating through the ground as it wants to continue doing due to the gravitational acceleration.

This is how you get no net force, but still feel the forces acting against gravity.

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  • $\begingroup$ I have a question related to your comment, but not really related to the OP question. You say: "When it is only gravity acting to pull you down (i.e. freefall) without any other forces, you cannot feel that gravity is acting on you. It feels the same as floating without any forces. What the toy car actually experiences is the upwards force from the ground pushing back on it." My question is: What is the reason why, even though we cannot feel the pulling of gravity, we are still "pulled" somehow to the ground? $\endgroup$ – AWanderingMind May 29 at 12:33
  • $\begingroup$ @AWanderingMind Usually it's called the equivalence principle en.wikipedia.org/wiki/Equivalence_principle Gravity actually seems to act more like a pseudoforce/inertial effect instead of an actual force being applied to the body, in general relativity. As far as human observation goes, that is also how we experience it. $\endgroup$ – JMac May 29 at 14:16
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Remember that force, velocity and acceleration are vector quantities. When you are talking about $\vec{F}, \vec{v}, \vec{a},$ you must also specify the direction of motion.

I am guessing that the direction of motion in your case is independent on the direction on which the gravity act upon.

So there is not net force on the direction of motion because the object is moving with constant velocity. And there is no net force on the up-down direction because the force of gravity is equaled by the normal action of the whatever the car is moving on (which is based on electric repulsion at the fundamental level).

$$F=ma=m\frac{d^2x}{dt^2}$$ if $v=\text{const}$ $$\frac{d}{dt}x=\text{const}$$ which means that: $$\frac{d^2x}{dt^2}=0$$

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