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enter image description here

In this picture if R is the component of P which is the reactionary force of mg, then according to the picture R is balancing mg in equilibrium. So my question is how can a component of a vector have same magnitude as that of a vector if its not perpendicular or parallel to it???

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  • $\begingroup$ Your question does seem to make sense. Could you define all of the vectors in your figure? I suspect that you have an inconsistent view of what each is with respect to the others or that one of the vectors does not belong on the figure. $\endgroup$ – Brick Sep 10 '18 at 15:56
  • $\begingroup$ the figure is about a object kept on a rough surface and when a force F is applied on it the Vector P which was the reactionary force of mg inclines to the left as F's direction is right. Now fs and R are the components of P and fs is static frictional force and R's has been described in the question. $\endgroup$ – Mad Dawg Sep 10 '18 at 16:00
  • $\begingroup$ R and mg are indeed parallel, they just have opposite directions. And R is a component of P, not of mg. $\endgroup$ – user190081 Sep 10 '18 at 16:04
  • $\begingroup$ thats what ive written man😐 $\endgroup$ – Mad Dawg Sep 10 '18 at 16:06
  • $\begingroup$ R is not parallel to P........ $\endgroup$ – Mad Dawg Sep 10 '18 at 16:06
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The component of your vector can have the same magnitude of another vector even if it is not in the same direction because of the principle of superposition: a vector could be seen as a sum of many other vectors along a system of axis. By that, you can thus decompose vectors along the lines that you want in their components.
So, now that you know that you can decompose a vector along the direction (that is "a line") that you want, you can decompose it along the same direction of another vector, to compare them. In our example, you decompose the vector P along the direction of vector $m\vec{g}$.

Parallel means that they lie on a same line (or at least parallel lines) ergo they have 0 angle between them, so you can compare them in a forces' problem like yours to see if there's equilibrium or acceleration.

EDIT: I added an image that make you see how you can use the superposition principle to remember that even if a vector is not on a precise direction, it has its component along that direction, that is the circled one that you don't have to forget

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Your figure is unconventionally made insofar as it suggests that all of the vectors represent independent forces whereas $P$, $R$, and $f_s$ are constrained such that two are the decomposition of the other.

If the block is in equilibrium, then the horizonal forces must balance, and, separately, the vertical forces must balance. That's a relationship between $F$ and $f_s$ and, separately, between $R$ and $mg$. Once you solve that, $P$ is determined.

I'm still not able to make sense of the question as you asked it. Solve the equations, and you have the vector forces. For each individual vector, the relationship between magnitude and direction depends on the axes you use for the decomposition. If you decompose along the direction of the vector, then one component is magnitude (and the others are 0), otherwise no component equals the magnitude. But that's got nothing apparent to do with the diagram since solving for the forces as described will not result in a situation that violates that expectation.

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You have either three forces acting (black in the diagram below) or four forces acting (not the forces labelled $P$.

enter image description here

Force $P$ is the sum of the normal reaction force $R$ and the frictional force $fs$.

The three force look as though they meet at a point $X$ and so it could be an equilibrium situation.

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  • $\begingroup$ yeah it is an equilibrium condition $\endgroup$ – Mad Dawg Sep 10 '18 at 16:14

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