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In Resnick, Halliday, Walker s' Principle of Physics , I found this question :

Two forces are acting on a block on a frictionless floor. On the left of the block $3~\text{N}$ & on the right of the block $5~\text{N}$ are acting. If a third horizontal force $\vec{F_3}$ also acts on the block, what are the magnitude and direction of $\vec{F_3}$ when the block is a)stationary and b)moving to the left with a constant speed of $5~\text{m/s}$ ?

What I have done: For being stationary, $2~\text{N}$ must be applied on the leftside of the block so as to make the net external force zero. In order to move with $5~\text{m/s}$ to the left, first we have to apply force greater than $5~\text{N}$ to the left side so as to accelerate it to the left. When the velocity reaches $5~\text{m/s}$,then we will weaken the force on the left to match the force on right ie. $5~\text{N}$ so that block moves at that required velocity.

My book's answer: Without explaining much, they wrote that in both the cases the force will be $2~\text{N}$ .

So, why did they tell that it was $2~\text{N}$ in the second case? What is the problem with my analysis? If it were $2~\text{N}$ , then the block will not move;at first I have to exert a greater force, then can only I reduce the force to let it move at that constant rate. So why am I wrong??

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    $\begingroup$ The wording asks about two cases: stationary and moving. It is silent about how they got to those states. You are told the block is moving. It's already reached it's speed an leveled off. $\endgroup$ – garyp Oct 31 '14 at 16:40
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Remember that a force causes acceleration. If you pull the block to the left with $5\mbox{N}$ when the velocity reaches $5\mbox{m/s}$, then the net force will be $2\mbox{N}$ towards the left. And the block will experience acceleration.

If you want steady speed, you need $0$ acceleration, and this is achieved by setting the net force to $0$.

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This is a common trip-up for first year physics students. Both part a) and part b) of your problem describe situations with no acceleration. A stationary object is not accelerating, but a moving object at constant velocity is not accelerating either. Therefore, Newton's Second Law tells us that in both situations, there is no net force on the block. $$\Sigma F=ma$$ and then $$\Sigma F=0$$ because $a=0$.

Physicists actually use different terms to describe the situations in part a) part b) in order to make the distinction between these two scenarios. The situation in part a) would be called a static equilibrium because the object is not moving. The situation in part b) would be called a dynamic equilibrium because even though there is no net force, the object is not stationary.

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  • $\begingroup$ Of course whether an object is "stationary" depends on your frame of reference, so the distinction you make in the final paragraph is a bit arbitrary. $\endgroup$ – Floris Oct 31 '14 at 16:54
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    $\begingroup$ Fair enough. Just pointing out there are times where the distinction is useful. $\endgroup$ – Sean Oct 31 '14 at 17:14
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As force acting on the right to the block is 5N and on left side it is 3N.Now if we want move the block to the left side with 5m/s then we have to apply a force just greater then 2N.so they may have taken a value 2.00000000000000001 which is almost equal to 2N in their consideration.

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