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I'm an A2 igsce math student and I'm taking mechanics for the first time in math this session. However, I was adviced to ask this question in the physics forum instead.

I know that what I'm going to say is wrong judging from my teacher's reaction when I asked him about this but unfortunately he didn't explain to me why it's wrong.

So, when we draw two forces on an object, we usually assume that they are applied at the same time but do we also assume that they were applied for a certain duration of time, like one second maybe?

I mean, how are we supposed to get the magnitude of the resultant of these two forces meanwhile we don't know the duration? To make my point clearer, for example, couldn't there be more than one force in the exact direction of the resultant force that would make our object/point reach the final speed it is supposed to reach after these two forces are applied for a certain time? — thinking about it now the examiner never mentions for how long two forces should be applied, but anyway.

But each force would need a certain amount of time, like a weaker force will need more time while a stronger one will take less time and so on. And all these forces are in the same direction of the resultant force of course.

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This question is brilliant, In the many institutions the time duration of two forces are assumed to be of the same duration to make things similar but anyway if you think of weak force and a strong force there is no problem to consider them being played on different durations but this might affect the answers.

Time is usually denoted by a scalar that is a number or a interval that is associated with the vector.

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  • $\begingroup$ Could you please give me an example to which how it can affect the answers? I'm trying to guess a scenerio myself but I feel like I'm getting nowhere :( $\endgroup$
    – Manar
    Commented Jan 17, 2021 at 12:07
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    $\begingroup$ Imagine rotating a cube and moving it simultaneously as you rotate it. If we think rotating as a weak force it means that it will last for a small duration while thinking moving as a strong force that is it will last for a long duration. Let the resultant force, vector be denoting the position of the cube. Now rotate and move the cube as the rotating is the weak force it will end soon but moving will last longer because we considered it as a strong force, now thing that vector has some angle and a fixed length. Would its angle be the same if you add little changes to rotating and moving.no. $\endgroup$
    – user285989
    Commented Jan 17, 2021 at 12:18
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One has not to confuse forces and effects of the forces. The resultant force's effect is the motion of the system to which the force has been applied. This motion is certainly different if a given force acts for a longer or shorter time. But the second Newton's law says that the acceleration at a given time $t$ is given once one knows the force at the same time. The instantaneous force information is encoded into a second-order differential equation whose solution automatically takes into account the time-span each individual force has been applied.

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  • $\begingroup$ So If I've a point and I apply to it at the same time for the same duration a horizontal force of $9N$ and a vertical force of $12N$, the resultant force should be $15N$ acting with an angle of $53.1°$ anticlockwise to the horizontal force of $9N$. And even if I increase that duration, say the first duration was $5$ seconds and now I make it $15$ seconds, the resultant force magnitude won't inscrease? Or will it? $\endgroup$
    – Manar
    Commented Jan 17, 2021 at 11:38
  • $\begingroup$ It will be the same if I'm allowed to say that the 15 N and the 9N and their resultant the 15 N happened at the same time and more importantly for the same period of time. If I'm allowed to say that I can make the resultant force happen for a shorter or longer period of time than its components will it then be the only case that the resultant will be different than 15 N right? $\endgroup$
    – Manar
    Commented Jan 17, 2021 at 12:00
  • $\begingroup$ @Manar, force, like position and velocity, is a function of time and we assign a value at each instant $t$. If you move, certainly your position will change if you keep the same velocity for different interval of time. Nevertheless, this still allows to speak about your position at a time $t$. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jan 17, 2021 at 12:32

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