Projection Using Dot Product Finding a Force (Boat Problem)

• Lebombo
In summary, the projection using dot product is a useful method for finding a force in a specific direction, as demonstrated in the boat problem where a 600 pound boat is sitting on a 30 degree inclined ramp. The unit vector v is used to represent the direction of the ramp and the force due to gravity is projected onto it to find the required force to keep the boat from rolling down the ramp. While any vector with a 30 degree angle would work, the unit vector is chosen for convenience. Additionally, the dot product of the force and distance vectors can be used to calculate work in this 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?

Lebombo said:

Homework Statement

-------------------------------------------------------------------------------------------
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 ProjvF is the projection of the force vector in the direction of v.

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.

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{<0,-600>.<100cos30, 100sin30>}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}$$\frac{<100cos30, 100sin30>}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}$ = $\frac{<0,-600>.<100\frac{\sqrt{3}}{2}, 100\frac{1}{2}>}{\sqrt{(100\frac{\sqrt{3}}{2})^{2} + (100\frac{1}{2})^{2}}}$$\frac{<100\frac{\sqrt{3}}{2}, 100\frac{1}{2}>}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}$$\frac{<0,-600>.<100\frac{\sqrt{3}}{2}, 100\frac{1}{2}>}{(\sqrt{(100\frac{\sqrt{3}}{2})^{2} + (100\frac{1}{2})^{2}})^{2}}$$<100\frac{\sqrt{3}}{2}, 100\frac{1}{2}>$= $\frac{30000}{10000}<100\frac{\sqrt{3}}{2},100\frac{1}{2}>$

= $\frac{30000}{10000}<100\frac{\sqrt{3}}{2},100\frac{1}{2}>$ = $3<100\frac{\sqrt{3}}{2},100\frac{1}{2}>$ = $<300\frac{\sqrt{3}}{2},300\frac{1}{2}>$

Lebombo said:
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.
Lebombo said:
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..."
Lebombo said:
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.
Lebombo said:
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{<0,-600>.<100cos30, 100sin30>}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}$$\frac{<100cos30, 100sin30>}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}$ = $\frac{<0,-600>.<100\frac{\sqrt{3}}{2}, 100\frac{1}{2}>}{\sqrt{(100\frac{\sqrt{3}}{2})^{2} + (100\frac{1}{2})^{2}}}$$\frac{<100\frac{\sqrt{3}}{2}, 100\frac{1}{2}>}{\sqrt{(100cos30)^{2} + (100sin30)^{2}}}$

$\frac{<0,-600>.<100\frac{\sqrt{3}}{2}, 100\frac{1}{2}>}{(\sqrt{(100\frac{\sqrt{3}}{2})^{2} + (100\frac{1}{2})^{2}})^{2}}$$<100\frac{\sqrt{3}}{2}, 100\frac{1}{2}>$= $\frac{30000}{10000}<100\frac{\sqrt{3}}{2},100\frac{1}{2}>$

= $\frac{30000}{10000}<100\frac{\sqrt{3}}{2},100\frac{1}{2}>$ = $3<100\frac{\sqrt{3}}{2},100\frac{1}{2}>$ = $<300\frac{\sqrt{3}}{2},300\frac{1}{2}>$

Mark44 said:
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).

Mark44 said:
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)

Last edited:

1. What is projection using dot product?

Projection using dot product is a mathematical technique used to find the component of a vector in a specific direction. It involves taking the dot product of the vector and a unit vector in the desired direction to calculate the magnitude of the projection.

2. How is projection using dot product used in finding a force in a boat problem?

In a boat problem, projection using dot product is used to find the component of the boat's force in the direction of the motion. This can help determine the necessary force to move the boat in a specific direction, or to keep it moving at a constant velocity.

3. Can projection using dot product be used to find the total force in a boat problem?

No, projection using dot product only calculates the component of a vector in a specific direction. To find the total force in a boat problem, other techniques such as vector addition may be needed.

4. What is the difference between projection using dot product and finding a force in a boat problem?

Projection using dot product is a mathematical technique, while finding a force in a boat problem involves applying physical principles and equations to solve a problem involving forces acting on a boat. Projection using dot product is just one aspect of finding a force in a boat problem.

5. What are some real-world applications of projection using dot product?

Projection using dot product has many applications, including in physics, engineering, and computer graphics. It can be used to calculate the work done by a force, the torque exerted on an object, or the intensity of light in a specific direction, among many others.

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