Vector Question

Question

The magnitude and direction exerted by two tugboats towing a ship are 1640 kilograms, N30°W, and 1260 kilograms, S60°W, respectively. Find the magnitude in kilograms, and the direction angle, in degrees, of the resultant force.

The magnitude of the resultant force is _________ kilograms

​(Do not round until the final answer. Then round to the nearest hundredth as​ needed.)

Answer 1

It may help to draw this on an x-y axis. The x-axis will be W on the left and E on the right. The y-axis will be up is N and down is S. First we must find the angles from zero. Draw the first vector 30° west of north, or LEFT of the upward y-axis. Measuring from zero (East or right x-axis), we have θ₁=120°.

The second vector is 60° southwest, or left of the downward x-axis. This angle becomes θ₂=-150°

now we need to find the i and j values of each vector. These are the forces acting in the x and y directions. These will help us add the vectors directionally.

v₁= v₁(cosθ₁)i + v₁(sinθ₁)j

v₁= 1640(cos120°)i + 1640(sin120°)j

v₂ = v₂(cosθ₂)i + v₂(sinθ₂)j

v₂= 1260(cos(-150°))i + 1260(sin(-150°))j

solve these and add the i and j values you will get

R = (v₁ + v₂)i + (v₁ + v₂)j

R = (1640(cos(120°)) + 1260(cos(-150°)))i + (1640(cos(120°)) + 1260(sin(-150°)))j

in order to find the resultant vector,

V_r = √((v₁+v₂)²i+(v₁+v₂)²j

so you would square both i and j values from the R equation and take the square root of their sum.

to find the angle, the formula is

tanθ = ((v₁ + v₂)i)/((v₁ + v₂)j)

I hope this all makes sense. I can do an online lesson and probably explain it better. Let me know if you need more help.

Answer 2

N 30° W means 30 degrees west of North, or 90 degrees. This is 90 + 30, or 120°.

S 60º W means 60 degrees less than South, or 270 - 60 = 210º.

The angle between these two angles is 210 √- 120, or 90º.

So we are working with forces at right angles to each other.

We can use the parallelogram law to compute the magnitude of the resultant vector:

R = √(A² + B² -2AB×cos(θ))

The cosine of 90 degrees is zero. If A = 1640 kg and B = 1260 kg, then

R = 2,068.14 kg

☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆

Now, there is no blank for it, but the problem asks for the direction of the resultant vector as well. If all you need is the magnitude of the resultant vector, the above is the quickest way to compute that.

Because the angles involved are special ones (multiples of 30, 60, and 90), there are some shortcuts that can be taken. Since those shortcuts do not apply to the general case, I did not employ them in the solution shown above. The above method works for all such problems, regardless of the angles and magnitudes of the vectors involved.

If the resultant angle is also required, there is a different way to approach such problems. First, decompose the component vectors into their x and y components.

Let the first vector be U, let its components be Vx and Vy, and let the corresponding angle of that vector be θ.

Let the second vector be U, let its components be Ux and Uy and the corresponding angle of that vector be φ.

Then

Vx = V*cos(θ) = 1640*cos(120º) = 1640*(-0.5)

= -820 kg

Vy= V*sin(θ) = 1640*sin(120º) = 1640*((1/2)*√3)

= 820√3 kg

Ux = U*cos(φ) = 1260*cos(210º) = 1260*(-0.5*√3)

= -630*√3 kg

Uy = U*sin(φ) = 1260*sin(210º) = 1260*(-0.5)

= -630 kg

We now have the x and y components of U and V.

The x and y components of R are:

〈Vx + Ux, Vx + Vy〉= 〈 -820 -630√3, 820√3 -630〉

If you take the square root of the squares of each of those components, you will get the same value as we did above.

To find the resultant angle, simply take the inverse tangent of y over x:

tan^-1[ (-820√3 -630) / (-820 -630√3) ] ≈ -22.47º

This is the angle clockwise from 180º, so

180º - 22.47º = 157.53º

Answer 3