# M1Watch

#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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10 years ago
#2
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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#3
Sorry, PQR and S are all forces acting on a 'particle'. Answer is 8.4N, 95 deg
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10 years ago
#4
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 and the angle will be

so for example for Q
and
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10 years ago
#5
(Original post by Mos Def)
Sorry, PQR and S are all forces acting on a 'particle'. Answer is 8.4N, 95 deg
Ah, okay.

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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#6
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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10 years ago
#7
For starters, P does give you something in the x-direction (2N)

You should get an x-resultant of
Spoiler:
Show

and a y-resultant of
Spoiler:
Show

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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10 years ago
#8
What board is this?
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10 years ago
#9
Edexcel
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