Find the magnitude and the direction of the resultant vector

[chwala]

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.

 

[MatinSAR]

Looks fine.

 

[Mark44]

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

$\left( \dfrac {3.154} {2.52} \right)$ $R = \sqrt {3.154^2 + 2.52^2} = \sqrt {16.3} = 4.03$ 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]

I'll follow up on the rest of the questions as a bonus treat for the weekend. Cheers guys.

 

[WWGD]

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

 

[chwala]

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.

 

[Mark44]

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.

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

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

$F_{Resultant}=\sqrt {4.8^2 + (-1.8)^2}=\sqrt {26.28} = 5.126$ 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.

$F = ma$
$a = \dfrac{5.126}{2} = 2.56$ 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.

???
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} ##)

 

[chwala]

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

 

[chwala]

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.