# Thread: force vector

1. ## force vector

I am currently studying vectors and i have a big confusion about something..

There are questions in the section such as:
A yacht is sailing at a speed of 34 knots and a wind encounters from the south east, find the final speed and direction of the yacht..

Now.. to work this question out you draw the triangle and use sin and cosine rules to work out the unknowns..

But i also get these questions and they require a different approach to solve:
A yacht is moving 10kmph in a south easterly direction and encounters a 3 kmph current from the north. Find the actual speed and direction of the yacht.

Now these questions are very similar but this question requires a different approach to solve using vectors which involves finding the vector in component form and adding the coordinates, then finding the scalar product to find the answer for the speed...

My problem is when i use one approach for a question (such as the second approach on the first question) i get an incorrect answer and vise versa

can anyone explain please..
thankyou

2. My problem is when i use one approach for a question (such as the second approach on the first question) i get an incorrect answer and vise versa

Umm. If done correctly, the same approach should be okay.

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A yacht is sailing at a speed of 34 knots and a wind encounters from the south east, find the final speed and direction of the yacht..

The problem is not complete. Which direction is the yacht sailing.

---------------------------
But i also get these questions and they require a different approach to solve:
A yacht is moving 10kmph in a south easterly direction and encounters a 3 kmph current from the north. Find the actual speed and direction of the yacht.

...this question requires a different approach to solve using vectors which involves finding the vector in component form and adding the coordinates, then finding the scalar product to find the answer for the speed...

The approach can be the same. You solve for the horizontal and vertical components of the vectors by using sine or cosine.

"south easterly" means the vector's direction is 45degrees south of east.
"from the north" means the vector's direction is vertically downward, or 90degrees south of east.

So collect all easterly components,
E = 10cos(45deg) +3cos(90deg)
E = 7.071 +0
E = 7.071 kph

And collect the southerly components,
S = 10sin(45deg) +3sin(90deg)
S = 7.071 +3
S = 10.071 kph

So, speed = sqrt(E^2 +S^2)
speed = sqrt[(7.071)^2 +(10.071)^2]
speed = sqrt(151.424)
speed = 12.305 kph ---------------answer.

And, direction of the yacht:
Based from the south axis,
tan(theta) = E/S = 7.071 /10.071 = 0.702
theta = arctan(0.702) = 35 degrees
Therefore, the yacht is actually sailing in the (South 35degrees East) direction. ---answer.

3. yes your answer is correct , but if you try solving the method involving working out sine (angle) which is measured anti clockwise from the x axis and plugging into the formula lengthcos(sineangle)i + lengthcos(sineangle)j you get a different answer

4. Originally Posted by whitestrat
yes your answer is correct , but if you try solving the method involving working out sine (angle) which is measured anti clockwise from the x axis and plugging into the formula lengthcos(sineangle)i + lengthcos(sineangle)j you get a different answer
Is that so?

Maybe. But I cannot check that because I do not know, nor do I understand, the formula: lengthcos(sineangle)i + lengthcos(sineangle)j

cosine of the sine of an angle?

Sorry.

5. Originally Posted by whitestrat
...
1.) A yacht is sailing at a speed of 34 knots and a wind encounters from the south east, find the final speed and direction of the yacht..

Now.. to work this question out you draw the triangle and use sin and cosine rules to work out the unknowns..

2.) But i also get these questions and they require a different approach to solve:
A yacht is moving 10kmph in a south easterly direction and encounters a 3 kmph current from the north. Find the actual speed and direction of the yacht.

...
Hello,

to #1:

to answer the problem I need the course of the boat and the speed of the wind.... Sorry

to #2:

Use a coordinate grid. The i-direction is pointing East and the j-direction is pointing North. Then the direction vector representing the course of the yacht is:
$\displaystyle \vec{y}=\left( \begin{array}{c}1\\ -1\end{array} \right)$
the length of the direction vector is $\displaystyle \sqrt{2}$. You need a factor s so that $\displaystyle s \cdot \sqrt{2} = 10~\Longrightarrow~s = 5 \cdot \sqrt{2}$. Therefore the vector describing direction and speed of the yacht is:

$\displaystyle \vec{v}=\left( \begin{array}{c}5 \cdot \sqrt{2}\\ -5 \cdot \sqrt{2}\end{array} \right)$

Doing the same considerations you'll get the vector describing speed and direction of the wind as:

$\displaystyle \vec{w}=\left( \begin{array}{c}0\\ -3\end{array} \right)$

the resulting vector is: $\displaystyle \vec{v} + \vec{w}= \left( \begin{array}{c}5 \cdot \sqrt{2}\\ -5 \cdot \sqrt{2}\end{array} \right) + \left( \begin{array}{c}0\\ -3\end{array} \right) = \left( \begin{array}{c}5 \cdot \sqrt{2}\\ -3-5 \cdot \sqrt{2}\end{array} \right)$

Therefore $\displaystyle | \vec{v} + \vec{w} | = 109+30 \cdot \sqrt{2} \approx 12.3055$

With the length of this vector and the values of the components you easily can calculate the angle.