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Mos Def
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#1
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#1
how do you find the resultant from a 'polygon of forces' ?

P = 2, p = 0deg
Q = 3, q = 40deg
R = 4, r = 130deg
S = 5, s = 150deg

Degrees are measured anticlockwise from the Ox line.

from this i drew a polygon, and managed to split it into 3 triangles, and then added up 3 resultants and they were wrong

+Rep for solution!
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nota bene
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What are PQRS? are they points such as (2,2), (3,3) etc.?

Although I don't see why adding the three triangles would be wrong...

What is the answer given in the book?
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Mos Def
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#3
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Sorry, PQR and S are all forces acting on a 'particle'. Answer is 8.4N, 95 deg
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thelostchild
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#4
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just split them up into components acting along the x and y axis, the x axis will be when the angle=0 and the y axis will be when the angle=90

add up all the different components the force will be \sqrt{x^2+y^2} and the angle will be \tan^{-1}(\frac{y}{x})

so for example for Q
Q_x=3\cos 40 and Q_y=3\sin 40
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nota bene
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(Original post by Mos Def)
Sorry, PQR and S are all forces acting on a 'particle'. Answer is 8.4N, 95 deg
Ah, okay.

Start with resolving for each force, and then use polar coordinates to get your result.

e.g. S_x=5cos(150)=-5cos(30)=5sqrt(3)/2 and S_y=5sin(150)=5sin(30)=5/2

What do you get for the other?

edit: thelostchild's answer is a bit more detailed...
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Mos Def
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#6
Report Thread starter 10 years ago
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ok i get all the components for Q R and S... can't get one for P as the angle is 0.

also i add the 3 up (rt3 + 4 + 2rt5) ....
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nota bene
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#7
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For starters, P does give you something in the x-direction (2N)

You should get an x-resultant of
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3\cos(40)+2+5\cos(150)+4\cos(130  )


and a y-resultant of
Spoiler:
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3\sin(40)+4\sin(130)+5\sin(150)


Now, use the relationships for x and y that thelostchild has given you.

edit: hmm, this does not give me the right answer
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EvenStevens
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#8
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What board is this?
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M.A.H
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#9
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Edexcel
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