Find the magnitude and the direction of the resultant vector

AI Thread Summary
The discussion focuses on calculating the magnitude and direction of resultant vectors in mechanics. The calculations for the resultant vector yield a magnitude of 4.03 N and a direction of 38.6° anticlockwise from the x-axis. Participants also address a second problem, leading to a resultant force of approximately 5.126 N with a direction of 20.6° in the x direction. There is a correction regarding the use of gravitational acceleration, with a note that the Cambridge textbook requires using g = 10 m/s². The conversation emphasizes the importance of accurate calculations and notation in vector analysis.
chwala
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Homework Statement
See attached
Relevant Equations
Mechanics
Going through this ( Revision) A salways your insights are quite helpful.

1710413541299.png



I would like to go through all these questions; i will start with (5),


##\left( \dfrac {x} {y} \right)## = ##\left( \dfrac {10 \cos 40^0} {10 \sin 40^0} \right)## + ##\left( \dfrac {4 \cos 150^0} {4\sin 150^0} \right)## + ##\left( \dfrac {6 \cos 260^0} {6 \sin 260^0} \right) ## = ##\left( \dfrac {7.66} {6.43} \right) ## + ##\left( \dfrac {-3.464} {2} \right)## + ##\left( \dfrac {-1.042} {-5.91} \right) ## =


##\left( \dfrac {3.154} {2.52} \right)##


##R = \sqrt {3.154^2 + 2.52^2} = \sqrt {16.3} = 4.03## N

For direction,

##\tan^{-1} \dfrac{2.52}{3.154} = tan^{-1} 0.79898 = 38.6^0## anticlockwise from the x-axis.
 
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Looks fine.
 
chwala said:
##\left( \dfrac {x} {y} \right)## = ##\left( \dfrac {10 \cos 40^0} {10 \sin 40^0} \right)## + ##\left( \dfrac {4 \cos 150^0} {4\sin 150^0} \right)## + ##\left( \dfrac {6 \cos 260^0} {6 \sin 260^0} \right) ## = ##\left( \dfrac {7.66} {6.43} \right) ## + ##\left( \dfrac {-3.464} {2} \right)## + ##\left( \dfrac {-1.042} {-5.91} \right) ## =
##\left( \dfrac {3.154} {2.52} \right)##
##R = \sqrt {3.154^2 + 2.52^2} = \sqrt {16.3} = 4.03## N

Expressing your results as fractions threw me off for a bit.
I would have used the LaTeX for matrices instead.
Here's your resultant as a column vector:
##\begin{bmatrix}3.154 \\ 2.52 \end{bmatrix}##
Click on what I wrote to see how I did this.
 
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I'll follow up on the rest of the questions as a bonus treat for the weekend. Cheers guys.
 
chwala said:
Homework Statement: See attached
Relevant Equations: Mechanics

Going through this ( Revision) A salways your insights are quite helpful.

View attachment 341764


I would like to go through all these questions; i will start with (5),

##\left( \dfrac {x} {y} \right)## = ##\left( \dfrac {10 \cos 40^0} {10 \sin 40^0} \right)## + ##\left( \dfrac {4 \cos 150^0} {4\sin 150^0} \right)## + ##\left( \dfrac {6 \cos 260^0} {6 \sin 260^0} \right) ## = ##\left( \dfrac {7.66} {6.43} \right) ## + ##\left( \dfrac {-3.464} {2} \right)## + ##\left( \dfrac {-1.042} {-5.91} \right) ## =


##\left( \dfrac {3.154} {2.52} \right)##
##R = \sqrt {3.154^2 + 2.52^2} = \sqrt {16.3} = 4.03## N
For direction,

##\tan^{-1} \dfrac{2.52}{3.154} = tan^{-1} 0.79898 = 38.6^0## anticlockwise from the x-axis.
Hey, I'm stuck in that labyrinth at the bottom of your post. Can you help me out??!!

Mentor note: Removed a bunch of HTML table tags
 
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For 6.
My lines are as follows,

##F_x = 24 + -19.2 = 4.8##

noting that ##g=-10## m/s^2.

##F_y = 7 + 14.8 + -20 = -1.8##

##F_{Resultant}=\sqrt {4.8^2 + (-1.8)^2}=\sqrt {26.28} = 5.126## to 3 decimal places.

##F = ma##

##a = \dfrac{5.126}{2} = 2.56## m/s^2.

For direction, i have ##\tan^{-1}\left[\dfrac{-1.8}{4.8}\right]=\tan^{-1} (-0.375)=- 20.6^0##.

Therefore the direction =##20.6^0## in the ##x## direction.
 
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chwala said:
For 6.
My lines are as follows,
##F_x = 24 + -19.2 = 4.8##
noting that ##g=-10## m/s^2.
That's a pretty rough approximation. A more precise value is -9.8 m/sec^2.
chwala said:
##F_y = 7 + 14.8 + -20 = -1.8##
I don't get a negative result, nor should you.
chwala said:
##F_{Resultant}=\sqrt {4.8^2 + (-1.8)^2}=\sqrt {26.28} = 5.126## to 3 decimal places.
It doesn't make sense to report a result to 3 decimal places when your value for g is so rough.
chwala said:
##F = ma##
##a = \dfrac{5.126}{2} = 2.56## m/s^2.
For direction, i have ##\tan^{-1}\left[\dfrac{-1.8}{4.8}\right]=\tan^{-1} (-0.375)=- 20.6^0##.
chwala said:
Therefore the direction =##20.6^0## in the ##x## direction.
???
Your previous result was -20.6°, which would mean that even with the engine running, the plane will not stay aloft very long. The resultant vector should have a direction that is positive relative to the x-axis, not one that is negative.

Also, in LaTeX you can write degrees using \circ rather than using 0 as an exponent. For example ##-20.6^{\circ}##. (Unrendered ##-20.6^{\circ} ##)
 
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Mark44 said:
That's a pretty rough approximation. A more precise value is -9.8 m/sec^2.

I don't get a negative result, nor should you.

It doesn't make sense to report a result to 3 decimal places when your value for g is so rough.


???
Your previous result was -20.6°, which would mean that even with the engine running, the plane will not stay aloft very long. The resultant vector should have a direction that is positive relative to the x-axis, not one that is negative.

Also, in LaTeX you can write degrees using \circ rather than using 0 as an exponent. For example ##-20.6^{\circ}##. (Unrendered ##-20.6^{\circ} ##)
Thanks @Mark44 , the cambridge textbook requires users to use ##g=10##m/s^2. Noted on the use of \circ ...also noted on my addition mistake...cheers
 
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chwala said:
For 6.
My lines are as follows,

##F_x = 24 + -19.2 = 4.8##

noting that ##g=-10## m/s^2.

##F_y = 7 + 14.8 + -20 = -1.8##

##F_{Resultant}=\sqrt {4.8^2 + (-1.8)^2}=\sqrt {26.28} = 5.126## to 3 decimal places.

##F = ma##

##a = \dfrac{5.126}{2} = 2.56## m/s^2.

For direction, i have ##\tan^{-1}\left[\dfrac{-1.8}{4.8}\right]=\tan^{-1} (-0.375)=- 20.6^0##.

Therefore the direction =##20.6^0## in the ##x## direction.
correct @Mark44

##F_y =7_+14.8 -20 = 1.8## a positive value. Typo error... cheers.
 
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