Projection Using Dot Product Finding a Force (Boat Problem)

[Lebombo]

"Projection Using Dot Product" "Finding a Force" (Boat Problem)

Homework Statement

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A 600 pound boat sits on a ramp inclined at 30 degrees. What force is required to keep the boat from rolling down the ramp?

Solution: Because the force due to gravity is vertical and downward, ou can represent the gravitational force by the vector F = -600j. To find the force required to keep the boat from rolling down the ramp, project F onto a unit vector v in the direction of the ramp, as follows.

v = cos30i + sin30j = \frac{\sqrt{3}}{2}i + \frac{1}{2}j


w_{1}= proj_{v}F = -300(\frac{\sqrt{3}}{2}i + \frac{1}{2}j)

Magnitude of force is 300.
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The question I have is why is vector v a unit vector? What determines that v should have a magnitude of 1?

 

[Mark44]

Homework Statement

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A 600 pound boat sits on a ramp inclined at 30 degrees. What force is required to keep the boat from rolling down the ramp?

Solution: Because the force due to gravity is vertical and downward, ou can represent the gravitational force by the vector F = -600j. To find the force required to keep the boat from rolling down the ramp, project F onto a unit vector v in the direction of the ramp, as follows.

v = cos30i + sin30j = \frac{\sqrt{3}}{2}i + \frac{1}{2}j


w_{1}= proj_{v}F = -300(\frac{\sqrt{3}}{2}i + \frac{1}{2}j)

Magnitude of force is 300.
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The question I have is why is vector v a unit vector? What determines that v should have a magnitude of 1?

Because it's convenient to have a unit vector. All you need is the direction. The notation Proj_vF is the projection of the force vector in the direction of v.

 

[Lebombo]

Okay, Mark, so the objective for including a vector v is to turn the scalor llwll into a vector w. That I understand. However it is which vector to choose that I am wondering about. From my understanding, I can choose a vector of any magnitude I choose, as long as it contains the correct direction. And I believe the reason I can choose a vector of any magnitude is because choosing any to multiply by any <b>v</b>/ll<b>v</b>ll is essentially multiplying the scalar by 1.

So the unit vector was chosen as the scalar simply for arithmetic convenience, but any vector of 30 degrees would produce the same answer as well. To test what the result of choosing a vector v of 30 degrees with a different magnitude, I chose a vector of magnitude 100. And it did in fact result in the same w as using the unit vector. Here is my work (please disregard the change in sign from negative to positive 300):w= \frac{F.v}{llvll}\frac{v}{llvll} =

\frac{&lt;0,-600&gt;.&lt;100cos30, 100sin30&gt;}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}\frac{&lt;100cos30, 100sin30&gt;}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}} = \frac{&lt;0,-600&gt;.&lt;100\frac{\sqrt{3}}{2}, 100\frac{1}{2}&gt;}{\sqrt{(100\frac{\sqrt{3}}{2})^{2} + (100\frac{1}{2})^{2}}}\frac{&lt;100\frac{\sqrt{3}}{2}, 100\frac{1}{2}&gt;}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}\frac{&lt;0,-600&gt;.&lt;100\frac{\sqrt{3}}{2}, 100\frac{1}{2}&gt;}{(\sqrt{(100\frac{\sqrt{3}}{2})^{2} + (100\frac{1}{2})^{2}})^{2}}&lt;100\frac{\sqrt{3}}{2}, 100\frac{1}{2}&gt;= \frac{30000}{10000}&lt;100\frac{\sqrt{3}}{2},100\frac{1}{2}&gt;

= \frac{30000}{10000}&lt;100\frac{\sqrt{3}}{2},100\frac{1}{2}&gt; = 3&lt;100\frac{\sqrt{3}}{2},100\frac{1}{2}&gt; = &lt;300\frac{\sqrt{3}}{2},300\frac{1}{2}&gt;

 

[Mark44]

Okay, Mark, so the objective for including a vector v is to turn the scalor llwll into a vector w. That I understand. However it is which vector to choose that I am wondering about. From my understanding, I can choose a vector of any magnitude I choose, as long as it contains the correct direction. And I believe the reason I can choose a vector of any magnitude is because choosing any to multiply by any v/llvll is essentially multiplying the scalar by 1.

I think you're confused by some of the terms. 1 is a scalar; v is a vector. If we multiply v by the scalar 1/||v||, we get a unit vector in the same direction as v.

So the unit vector was chosen as the scalar simply for arithmetic convenience,

Change this to "So the unit vector was chosen [STRIKE]as the scalar[/STRIKE] simply for [STRIKE]arithmetic[/STRIKE] convenience..."

but any vector of 30 degrees would produce the same answer as well.

Yes, any vector that makes a 30° with the horizontal will work. Since all we're interested in is the direction, it's convenient to use a unit vector that points in the right direction.

To test what the result of choosing a vector v of 30 degrees with a different magnitude, I chose a vector of magnitude 100. And it did in fact result in the same w as using the unit vector. Here is my work (please disregard the change in sign from negative to positive 300):


w= \frac{F.v}{llvll}\frac{v}{llvll} =

\frac{&lt;0,-600&gt;.&lt;100cos30, 100sin30&gt;}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}\frac{&lt;100cos30, 100sin30&gt;}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}} = \frac{&lt;0,-600&gt;.&lt;100\frac{\sqrt{3}}{2}, 100\frac{1}{2}&gt;}{\sqrt{(100\frac{\sqrt{3}}{2})^{2} + (100\frac{1}{2})^{2}}}\frac{&lt;100\frac{\sqrt{3}}{2}, 100\frac{1}{2}&gt;}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}


\frac{&lt;0,-600&gt;.&lt;100\frac{\sqrt{3}}{2}, 100\frac{1}{2}&gt;}{(\sqrt{(100\frac{\sqrt{3}}{2})^{2} + (100\frac{1}{2})^{2}})^{2}}&lt;100\frac{\sqrt{3}}{2}, 100\frac{1}{2}&gt;= \frac{30000}{10000}&lt;100\frac{\sqrt{3}}{2},100\frac{1}{2}&gt;

= \frac{30000}{10000}&lt;100\frac{\sqrt{3}}{2},100\frac{1}{2}&gt; = 3&lt;100\frac{\sqrt{3}}{2},100\frac{1}{2}&gt; = &lt;300\frac{\sqrt{3}}{2},300\frac{1}{2}&gt;

 

[Lebombo]

I think you're confused by some of the terms. 1 is a scalar; v is a vector. If we multiply v by the scalar 1/||v||, we get a unit vector in the same direction as v.

Okay, I think I am organizing my understanding of the terms better here:

w= \frac{F.v}{llvll}\frac{v}{llvll} =

\frac{F.v}{llvll} - is the scalar.

\frac{v}{llvll} - is the vector (unit vector).

we can use a vector of magnitude 1, 10, 20 or any magnitude because when the numerator and denominator are the same, they produce a unit vector of magnitude 1 anyway. When the denominator of the unit vector v/llvll is moved into the scalar to produce the familiar formula:

w= (\frac{F.v}{llvll^{2}})

the unit vector of magnitude 1 still exists, so even though you only see that you are multiplying by a vector (which is actually the numerator of the unit vector).

Yes, any vector that makes a 30° with the horizontal will work. Since all we're interested in is the direction, it's convenient to use a unit vector that points in the right direction.

Here's a follow up question. So we are calculating Force in this problem. And work is supposed to be the dot product of the Force vector and the Distance vector. So suppose the boat travels some distance, what exactly gets dotted with what?

Suppose the distance is some magnitude 10 in the horizontal direction. Should the dot product for work be:

(\frac{F.v}{llvll^{2}})v dotted with <10,0> ?

(note; I don't know how to do the dot symbol, any tips on that would be appreciated)