Find the magnitude and the direction of the resultant vector

• chwala
chwala
Gold Member
Homework Statement
See attached
Relevant Equations
Mechanics
Going through this ( Revision) A salways your insights are quite helpful.

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.

Last edited by a moderator:
Looks fine.

chwala
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.

chwala and SammyS
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

Last edited by a moderator:
MatinSAR, chwala and SammyS
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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MatinSAR and chwala
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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