Statics of a particleWatch

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Thread starter 14 years ago
#1
Two forces P and Q act on a particle at the origin O. Force P acts along the 0x, Q makes an angle with 0x. Calculate the Magnitude of the resultant force and the angle it makes with 0x when:

P = 4N, Q=3N feta =60

We are given the angle 60, according to my drawing this angle is outside the right angled triangle, and so therefore to find the inner angle that is opposite to the angle we are eventually to find - I did this:

180 - 60 = 120

We are given two lengths, so in order to find R we must use the cosine rule:

R^2=b^2+c^2 - 2bc cos A

R^2 = 3^2 + 4^2 - 2(3)(4) * cos120
= 9+ 16 - 24 * cos120
R = sqrt37
=6.08

According to my diagram, it would be wise to the sine rule as this traingle is not a <ABC triangle.

R / sin120 = 3/sin feta
sin feta therefore equals = 3 sin120 / sqrt37

which is wrong the angle should be 25.3

Thank you v much!
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14 years ago
#2
What are you trying to do? I'm not sure I follow.
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14 years ago
#3
It's theta, not feta. That's a type of cheese
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Thread starter 14 years ago
#4
(Original post by juno_the)
It's theta, not feta. That's a type of cheese
lol 0
Thread starter 14 years ago
#5
(Original post by Nima)
If you tell me whether P and Q act towards or away from the origin, then I can answer the question I think.
It doesn't say, but my feeling is that P acts along the x axis, whereas Q acts along the y axis and makes an angle of 60 on the x axis/
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14 years ago
#6
The question already gives you the angle. Why are you trying to find another one?
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Thread starter 14 years ago
#7
(Original post by dvs)
The question already gives you the angle. Why are you trying to find another one?
I am trying to find the inner angle of the triangle , I will draw picture of what I think the question looks like, bare with me for a sec.
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14 years ago
#8
Ohhh nevermind. I didn't read the question carefully, sorry!
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Thread starter 14 years ago
#9
(Original post by DOJO)
I am trying to find the inner angle of the triangle , I will draw picture of what I think the question looks like, bare with me for a sec.
This is my interpretation of the Question
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14 years ago
#10
The resultant force equals sqrt N. Apply the sine rule to get:
(sinT)/3 = (sin120)/sqrt [120 is from 180-60 -- draw the diagram and complete a parallelogram]

sinT = 3(sin120)/sqrt
T = 25.3
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Thread starter 14 years ago
#11
(Original post by dvs)
The resultant force equals sqrt N. Apply the sine rule to get:
(sinT)/3 = (sin120)/sqrt [120 is from 180-60 -- draw the diagram and complete a parallelogram]

sinT = 3(sin120)/sqrt
T = 25.3
I am so sure I did that, I don't know why I keep getting 0.427121098
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14 years ago
#12
Here's my interpretation. (I appologize for the crappy drawing.)
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Thread starter 14 years ago
#13
(Original post by dvs)
Here's my interpretation. (I appologize for the crappy drawing.)
Np, but it looks like our methods are both correct. I have no clue that when I do

sinT = 3(sin120)/sqrt
T = 25.3

It comes out as 0.427121098 and not 25.3. My calc is set in degrees mode.
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14 years ago
#14
(Original post by DOJO)
It comes out as 0.427121098 and not 25.3. My calc is set in degrees mode.
I really should start reading the entire post before replying...

First time, I read that the question only wanted you to find the resultant force, so I got confused when I saw you trying to find an angle. Second time, I thought you didn't know how to do the question. And now, I finally know what's wrong... You're finding sinT, and not T. Try finding arcsin(0.427121098) (inverse sine). If your calculator's one of those scientific Casios, press shift and sin, then = (after you get 0.427121098 on the screen).
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Thread starter 14 years ago
#15
(Original post by dvs)
I really should start reading the entire post before replying...

First time, I read that the question only wanted you to find the resultant force, so I got confused when I saw you trying to find an angle. Second time, I thought you didn't know how to do the question. And now, I finally know what's wrong... You're finding sinT, and not T. Try finding arcsin(0.427121098) (inverse sine). If your calculator's one of those scientific Casios, press shift and sin, then = (after you get 0.427121098 on the screen).
You are a genius, I really was wondering what I was doing wrong as I was pretty sure my methodology for that question was correct. Thanks!
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