# Solve Basic Vector Problem: Resultant & Angle \alpha

• logan3
In summary: The book solution is probably using the absolute value of θ, which is why they get 69°. Another way to think about it is that the angle with respect to the y-axis is either 21° or 69°, depending on which side of the y-axis the resultant falls on.
logan3

## Homework Statement

In a plane, find the resultant of a 300-N force at $30^o$ and a -250-N force at $90^o$. Also, find the angle $\alpha$ between the resultant and the y axis.

## Homework Equations

$x = F cos \theta$
$y = F sin \theta$
$r^2 = x^2 + y^2$
$\theta = tan^{-1} (\frac {y} {x})$

## The Attempt at a Solution

$x_1 = 300 cos 30^o = 259.80, x_2 = -250 cos 90^o = 0, x_{TOT} = 259.80$
$y_1 = 300 sin 30^o = 150, y_2 = -250 sin 90^o = -250, y_{TOT} = -100$
$r = \sqrt {259.80^2 + (-100)^2} = 278.38 \sim 280$

$\theta = tan^{-1} (\frac {259.80} {-100}) = -68.947^o \sim -69^o$
Since this is the angle between the x-axis and resultant, then the angle $\alpha$ between the resultant and the y-axis must be $-90 - (-68.947^o) = -21.053^o \sim -21^o$. However, the book solution for the angle $\alpha$ is given as $69^o$. Why?

Thank-you

logan3 said:

## Homework Statement

In a plane, find the resultant of a 300-N force at $30^o$ and a -250-N force at $90^o$. Also, find the angle $\alpha$ between the resultant and the y axis.

## Homework Equations

$x = F cos \theta$
$y = F sin \theta$
$r^2 = x^2 + y^2$
$\theta = tan^{-1} (\frac {y} {x})$

## The Attempt at a Solution

$x_1 = 300 cos 30^o = 259.80, x_2 = -250 cos 90^o = 0, x_{TOT} = 259.80$
$y_1 = 300 sin 30^o = 150, y_2 = -250 sin 90^o = -250, y_{TOT} = -100$
$r = \sqrt {259.80^2 + (-100)^2} = 278.38 \sim 280$

$\theta = tan^{-1} (\frac {259.80} {-100}) = -68.947^o \sim -69^o$
Since this is the angle between the x-axis and resultant, then the angle $\alpha$ between the resultant and the y-axis must be $-90 - (-68.947^o) = -21.053^o \sim -21^o$. However, the book solution for the angle $\alpha$ is given as $69^o$. Why?

Thank-you

Nice work. I've bolded the part where you transposed the y & x components when calculating theta...

logan3
Is the problem that you mixed up your terms in the formula for θ.
I think the formula for θ, the angle with respect to the x-axis is: tan -1( y/x).
It looks to me like you have x and y transposed, as y=-100 and x = 259.8, so should the expression be:
θ= tan -1(-100/259.8)?
If i use that expression I get θ ≈ -21° so with respect to the y-axis ∠ = -69.

## What is a vector?

A vector is a mathematical quantity that has both magnitude (size) and direction. It can be represented graphically by an arrow, with the length of the arrow representing its magnitude and the direction of the arrow representing its direction.

## What is a resultant vector?

A resultant vector is the combination of two or more vectors. It represents the overall effect of the individual vectors and is found by adding or subtracting the individual vectors using vector algebra.

## How do you find the magnitude of a resultant vector?

To find the magnitude of a resultant vector, you can use the Pythagorean theorem, which states that the square of the hypotenuse (longest side) of a right triangle is equal to the sum of the squares of the other two sides. In the case of vectors, the magnitude of the resultant vector is equal to the square root of the sum of the squares of the individual vectors' magnitudes.

## How do you find the direction (angle) of a resultant vector?

To find the direction of a resultant vector, you can use trigonometric functions such as sine, cosine, and tangent. The direction (angle) of the resultant vector can be found by taking the inverse tangent of the ratio of the vertical component to the horizontal component of the vector.

## Can the angle \alpha be negative?

Yes, the angle \alpha can be negative. This indicates that the resultant vector is in the opposite direction of the reference angle. In vector notation, a negative angle is represented by a negative sign in front of the angle value.

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