Mechanics

- TrojanPoem

Mechanics

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- TrojanPoem

|dw:1444160889035:dw|

- TrojanPoem

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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## More answers

- TrojanPoem

Resultant = 180

- TrojanPoem

The direction of the resultant is the same as the force 120

- TrojanPoem

@phi

- phi

What two vectors are we adding to produce the resultant ?

- TrojanPoem

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 ?

- TrojanPoem

The triangle is drawn

- TrojanPoem

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 ?

- TrojanPoem

|dw:1444162127993:dw|

- TrojanPoem

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?

- TrojanPoem

Yeah

- TrojanPoem

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

- TrojanPoem

@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

- TrojanPoem

@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.

- TrojanPoem

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.

- TrojanPoem

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

- TrojanPoem

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!)

- TrojanPoem

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.

- TrojanPoem

If we used direction beta, we would use the anti-clockwise angle , right ?

- TrojanPoem

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
\[ F^2 \cos^2 A + F^2 \sin^2 A =(54.73)^2 + (-82.23)^2 \\
F^2(\cos^2 A +\sin^2 A)=(54.73)^2 + (-82.23)^2 \\
F^2 = (54.73)^2 + (-82.23)^2 \\
F= \sqrt{(54.73)^2 + (-82.23)^2}\]

- TrojanPoem

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

- TrojanPoem

Yeah, I took " rectangular to polar" before, I think in imaginary numbers.

- phi

are you posting a different question ?

- TrojanPoem

Yeah.

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