I'm supposed to solve this using vectors in the form of v = |v|[(cos X)i + (sin X)j], but I only know ...?

Question

... how to solve it using the law of cosines! Can someone please help me please????

It says:

A boat heads 35 degrees (does not say from where), propelled by a force of 750 lbs. A wind FROM 320 degrees exerts a force of 150 lbs on the boar. How large is the resultant force, and in which direction is the boat moving?

I have no idea how to set this up or anything!!! Can someone please walk me through this?! I would appreciate it so much! Thank you so much and I will pick a best answer TODAY!!!!!!!!!!!!


Also, can anyone tell me what type of math class you would find this in (i.e. precalc, calc 1, calc 2, linear algebra, etc.)?

Answer 1

Ok, what you need to do is a vector sum. What you have is two vectors. I'm going to assume that the boat is moving at 35 degrees from the horizontal (in a North-East direction, if you imagine the coordinate plane as a compass). Your two vectors are:

(a) The boat's direction vector, which is the force of 750 in the direction of 35 degrees:

a= 750

(b) The wind's force vector, which is the force of 150, coming in from 320 degrees. This part is a bit tricky. Imagine an arrow (the force vector) coming out from the origin at the angle of 320. We need the opposite of this arrow (or the negative), so that it points TOWARD the origin (which is where the boat is). So we have:

b= 150<-cos320 i, -sin320 j>
From here, all you need to do is add the vectors (you just add them component-wise). Distribute the magnitudes (750 and 150) first, to make it easier:

a= <750cos35 i, 750sin35 j> = <614.36 i, 430.18 j>
b= <-150cos320 i, -150sin320 j> = <-114.91 i, 96.42 j>

Thus,
a + b = <614.36 - 114.91 i, 430.18 + 96.42 j>
a + b = <499.45, 526.60>

Now, this resultant vector combines the magnitude of the resultant force with the direction, so we need to break it apart. The resultant vector, symbolically, is as you wrote in the question:

a + b =

You can set the components equal to each other now, and you have two equations and two unknowns:

vcosX = 499.45
vsinX = 526.60

To solve this, square both sides of each equation, then add them:

(v^2)(cosX)^2 = 249450
(v^2)(sinX)^2 = 277308

(v^2)(cosX)^2 + (v^2)(sinX)^2 = 526758

You have a sin^2 and a cos^2, which drop out to become 1 by the trig identity, leaving you with:

v^2 = 526758
=>
v = 725.78

That is your resultant force. Plug it back into one of the earlier equations to get your angle:

(725.78)cosX = 499.45
cosX = .6882
X = 46.5 degrees from the horizontal

You can write your answer as a force of 725.78 lbs at 46.5 degrees NE (northeast being quadrant 1).

Hope this helped!

Oh, and this kind of question would be found in class dealing with vectors (obviously). At the high school level, this is usually in BC calc (vectors making up the 'C' component). In college, this is the beginnings of multivariable calculus. It also comes up in physics.

Answer 2

I assume that "heads 35°" means "has a compass heading of 35°". 0° is north; 35° is 35° east of north (northeast is 45°). The wind coming FROM 320° means it is blowing TOWARD 140° (a bit south of southeast).

Assigning a coordinate system is arbitrary. For the work below I will let North coincide with the positive y-axis. So the boat's force vector makes a 35°angle with the positive y-axis and so a 55° angle with the positive x-axis. The magnitude of the boat's force vector is 750 (lbs). The boat force vector is

750 cos(55°)i + 750 sin(55°)j

The wind's force vector makes an angle of 140° clockwise with the positive y-axis; so it makes an angle of -50° with the x-axis. The magnitude of the wind force is 150 (lbs). The wind force vector is

150 cos(-50°)i + 150 sin(-50°)j

Since cos(-a) = cos(a) and sin(-a) = -sin(a), this may be written

150 cos(50°)i - 150 sin(50°)j

The net force is the vector sum:

F = [750 cos(55°) + 150 cos(50°)] i + [750 sin(55°) - 150 sin(50°)] j

which is approximately 526.60 i + 499.46 j

The magnitude of F is the square root of the sum of the squares of the components. The angle that F makes with the positive x-axis is

inverse tangent of (j-component of F)/(i-component of F),

which is about 43.48°. This angle is measured from the positive x-axis. In terms of compass heading, it is (90° - 43.48°) = 46.52°

I strongly urge you to draw a fairly accurate diagram of this setup, to refer to while working through this. It really helps.

I don't know what type of math class you would find this in. I think most people see vectors before calculus.

Answer 3

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