Finding the resultant of two vector of an oblique triangle

[ElizabethC7]

Homework Statement

A car travels 20.0 km due north and then 35.0 km in a direction 60.0° west of north. Using a graph, find the magnitude and direction of a single vector that gives the net effect of the car's trip. This vector is called the car's resultant displacement.

I am studying for a physics test, and this is a problem out of my textbook. The author gives the answer of the resultant to be 48 km, and the angle β, the angle between vector A and the resultant, to be 39° west of north.

Homework Equations

From the book, because I missed class the day we went over this, I have learned that to find the resultant one simply adds vector A to vector B.

R= A + B

Also, from different physics web sites, some people have suggested the pythagorean theorem; however, I believe that only works with right triangles, not oblique.

R²= A² + B²

Lastly, I came across this equation.

c²= a² + b² - 2abcos(c)

The Attempt at a Solution

I have tried all 3 equations.

1) R= 20 + 35= 55

2) R²=20² +35²
R²= 1625
R= (sqrt 1625)
R= 40.3112887

3) R²= 20² + 35² - 2(20)(35)cos(60)
R²= 1625-1400cos(60)
R²=1625-933.71
R²= 691.29
R= 26.2923943


Obviously, none of these solutions match the one given by the author. How am I supposed to work this problem to find the right solution? As far as finding angle β, I have no idea where to start

 

[gneill]

Add vectors by adding their components. The result is the components of, you guessed it, the resultant! So first break down your given vectors into their components.

The two components of your resultant will be at right angles to each other, just as in a Pythagorean triangle. So use the appropriate formula to find the magnitude of that vector from its components.

Also, look at your graph of the setup to determine how you might find the requested angle of the resultant.

 

[ElizabethC7]

Thank you for your help, but could you explain how I am to find vector components?

 

[gneill]

Thank you for your help, but could you explain how I am to find vector components?

Draw a diagram of a vector on a set of Cartesian coordinates. The components of the vector are the perpendicular projections of the vector on the coordinate axes.

If the vector projects from the origin to a point (x,y), then x and y are also the magnitudes of the components. Clearly, then, given a vector of length R that makes some angle θ with the x-axis, the components can be found using the basic trig formulas for a right triangle.

 

[rwisz]

.

To find the resultant of two vectors in an oblique triangle, you can use the law of cosines: c^2 = a^2 + b^2 - 2abcos(C), where c is the length of the resultant, a and b are the lengths of the two vectors, and C is the angle between them. In this case, c represents the resultant displacement of the car.

Using this formula, we can plug in the values given in the problem: c^2 = (20 km)^2 + (35 km)^2 - 2(20 km)(35 km)cos(60°) = 400 + 1225 - 700(0.5) = 625. Taking the square root of both sides, we get c = 25 km.

Therefore, the magnitude of the resultant displacement is 25 km. To find the direction, we can use the law of sines: sin(A)/a = sin(B)/b = sin(C)/c, where A and B are the angles opposite sides a and b, respectively. In this case, we know the lengths of sides a, b, and c, and we can solve for the angle opposite side c, which is the angle between the two given vectors.

sin(C)/c = sin(60°)/35 km = sin(β)/25 km. Solving for β, we get sin(β) = (25 km)(sin(60°))/35 km = 0.714. Taking the inverse sine of both sides, we get β = 45.53° west of north.

Therefore, the resultant displacement of the car is 25 km at an angle of 45.53° west of north. This is the vector that represents the net effect of the car's trip. I hope this helps in your preparation for your physics test.