Mechanics

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|dw:1444160889035:dw|
F_2 = 186.3 / 98.8 / 47.7 / 72.2 beta = 303.6 / 123.6 / 236.4 / 56.4
  • phi
what does 120 N R= 180 N mean ?

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Other answers:

Resultant = 180
The direction of the resultant is the same as the force 120
  • phi
What two vectors are we adding to produce the resultant ?
It's the result of adding all the forces
  • phi
so we are adding 4 vectors? in terms of mag, angle <170, 50> <200, 180> <120, ??> do we have to find the angle for that last vector ?
The triangle is drawn
In mechanics they mostly draw the triangle on the line like this:|dw:1444162046880:dw|
  • phi
ok, so we need the angle. what is it ?
|dw:1444162127993:dw|
We are looking for beta and F_2
  • phi
ok, here are the 4 vectors <170, 50> <200, 180> <120, 126.87> and the result is <180, 126.87> correct?
Yeah
Will you use -beta ? as anti clockwise ?
  • phi
I would write down each vector in components and add them up.
  • phi
Here are the details change to rectangular coords using where are polar coords we get <170, 50> -> <109.27, 130.23> -> < F cos A, F sin A> <200, 180> -> <-200, 0 > <120, 126.87> -> < -72, 96 > --------------------------- <180, 126.87> -> <-108, 144> add up the 4 vectors, by adding the their x components and their y components, and set equal to the resultant's components we get F cos A - 162.72= -108 F sin A + 226.23= 144 or F cos A = 54.73 F sin A = -82.23 vector F in rectangular coords is <54.73, -82.23> in polar coords A = atan( -82.23/54.73) = -56.35 deg F= sqr( 54.73^2 + (-82.23)^2 ) = 98.78
@phi , |dw:1444238623491:dw| 170Cos50 + F_2Cosbeta - 120 * 3/5 - 200 = -3/5 * 180 ( resultant component in x-axis) F_2CosBeta = 54.73 170Sin50 + 120 * 4/5 - F_2SinBeta = 180 * 4/5 ( resultant component in y - axis) F_2SinBeta = 82.23 TanBeta = 82.23/54.73 <(beta) = 360- 56 21 = 303.6 F = 82.23/sin303.6 = -98.8 F = 54.73/ cos303.6 = 98.8 OK, where is the mistake ? I keep getting -ve , +ve values for the same force :/ help !
  • phi
You have **170Sin50 + 120 * 4/5 - F_2SinBeta = 180 * 4/5*** I would use 170Sin50 + 120 * 4/5 + F_2SinBeta = 180 * 4/5 so you get F_2 Sin Beta= -82.23
@phi , But why did you change the sign to +Ve ? Sin(360- beta) = -Sinbeta ? If so why can't I use the normal beta, I am breaking the force into its components.
I was told to use the anti-clockwise angle when giving out answers not breaking the force.
  • phi
My thought to use -beta is wrong. We should use Beta and solve for it. we will get 303 which also can be written as -57. The important thing is r cos beta and r sin beta give the correct magnitude and direction of each component. from the diagram we know x is positive and y is negative.
So the negative here points to the direction of the force component ( -ve y axis ) ? so it's not wrong that the force gives two signs ?
  • phi
the magnitude F2 should always be positive notice if we use -82 , we will get a positive value for F2 using -82/sin 303
F_2SinBeta = 82.23 I have 82 as positive, what did I do wrong ?
  • phi
when you solve for F_2 and Beta, use the terms r cos beta and r sin beta (you used -r sin beta for the y value)
  • phi
in other words, let's not over-think this. the components are always r cos A , r sin A (no -A angles!)
I used - r sin beta as I assumed that the resultant direction is on the +ve y axis so anything in +ve -y axis = + , anything in -ve y axis = -ve
  • phi
no, we vectors that have magnitude r and direction beta. we should just use r and beta.
If we used direction beta, we would use the anti-clockwise angle , right ?
I think using your method of solving them would be easier.
  • phi
if we convert from polar to rectangular we get <170, 50> -> <109.27, 130.23> -> < F cos A, F sin A> <200, 180> -> <-200, 0 > <120, 126.87> -> < -72, 96 > --------------------------- <180, 126.87> -> <-108, 144> add them up we get F cos A - 162.72= -108 F sin A + 226.23= 144 F cos A = 54.73 F sin A = -82.23 if we divide the bottom equation by the top we get \[ \frac{F \sin A}{F \cos A}= \frac{ -82.23}{54.73} \\ \tan A = \frac{ -82.23}{54.73} \] notice the y value is neg and the x value is pos, that means we are in the 4th quadrant. if we square both equations we get \[ F^2 \cos^2 A = (54.73)^2 \\ F^2 \sin^2 A = (-82.23)^2 \] add them to get \[ F^2 \cos^2 A + F^2 \sin^2 A =(54.73)^2 + (-82.23)^2 \\ F^2(\cos^2 A +\sin^2 A)=(54.73)^2 + (-82.23)^2 \\ F^2 = (54.73)^2 + (-82.23)^2 \\ F= \sqrt{(54.73)^2 + (-82.23)^2}\]
Squaring both sides, really ingenious idea. Cos^2A + Sin^2A = 1 That way we can find the force first thanks :)
  • phi
that is the background on how to change from rectangular to polar (x,y) A= atan(y/x) r= sqr(x^2 + y^2) you can also do r= x/cos A or r= x/sin A
Yeah, I took " rectangular to polar" before, I think in imaginary numbers.
  • phi
are you posting a different question ?
Yeah.

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