# Statics- Resolving forces

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#1
This link contains an image of a force diagram:

https://imgur.com/a/TziH0lP

I need to find the values of p and q. I know the general idea of this: resolve parallel to and perpendicular to the plane, equate resultant force to zero, and rearrange for p and q. That's not an issue. I know it sounds odd, but the horizontal force of 2N here is the only one giving me trouble. I don't really visualise the angle i'm to work with here. This has happened before on another occasion, it's particularly the horizontal forces going into the slope. Could someone annotate the image i've attached and show me what i'm supposed to work with here?

Thanks
Last edited by Illidan2; 2 years ago
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2 years ago
#2
(Original post by Illidan2)
This link contains an image of a force diagram:

https://imgur.com/a/TziH0lP

I need to find the values of p and q. I know the general idea of this: resolve parallel to and perpendicular to the plane, equate resultant force to zero, and rearrange for p and q. That's not an issue. I know it sounds odd, but the horizontal force of 2N here is the only one giving me trouble. I don't really visualise the angle i'm to work with here. This has happened before on another occasion, it's particularly the horizontal forces going into the slope. Could someone annotate the image i've attached and show me what i'm supposed to work with here?

Thanks
Z-angles (i.e. alternate angles) from GCSE maths.

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#3
(Original post by RDKGames)
Z-angles (i.e. alternate angles) from GCSE maths.

That thought did occur to me yes, meaning the angle contained between the force 2N and the plane would be 60 degrees. But then how would I resolve the 2N force? into a line going up the plane(parallel to) and another going perpendicular to the plane? I can't really visualise what it would look like, it sounds strange I know, but it's the only force on the diagram that I don't really understand what it would look like if it was split into its components.
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2 years ago
#4
(Original post by Illidan2)
That thought did occur to me yes, meaning the angle contained between the force 2N and the plane would be 60 degrees. But then how would I resolve the 2N force? into a line going up the plane(parallel to) and another going perpendicular to the plane? I can't really visualise what it would look like, it sounds strange I know, but it's the only force on the diagram that I don't really understand what it would look like if it was split into its components.
The way I like to visualise it is to mark down the starting point (blue) and the ending point (red) of the force I am resolving, then just draw a path made up of the perpendicular and parallel components connecting the two. You know an angle and the hypotenuse, so working out those components is no trouble.

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#5
(Original post by RDKGames)
The way I like to visualise it is to mark down the starting point (blue) and the ending point (red) of the force I am resolving, then just draw a path made up of the perpendicular and parallel components connecting the two. You know an angle and the hypotenuse, so working out those components is no trouble.

Yeah, your diagram is consistent with what I imagined would be the way the components acted. So if we were resolving up the plane, for example, we'd have:

Q+2cos(60)-6sin(60)=0
Q=4.20N(3 s.f)

Right?
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2 years ago
#6
(Original post by Illidan2)
Yeah, your diagram is consistent with what I imagined would be the way the components acted. So if we were resolving up the plane, for example, we'd have:

Q+2cos(60)-6sin(60)=0
Q=4.20N(3 s.f)

Right?
Yep
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#7
(Original post by RDKGames)
Yep
I see thank you very much for your help on this
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