Vector decomposition - gravity

In summary: We can see, by drawing a scale diagram, that ##\vec W_2## (the component of weight normal to the slope) will be equal in magnitude and opposite in direction to ##\vec W_1##. In other words, ##\vec W_2 = -\vec W_1##. So we don’t need to do any further calculations. We just write down:##\vec W_2 = <\frac {5√3}{2}, \frac 5 2>##Note: the angle of the slope depends on whether we choose the x-axis to point to the right (as I did) or to the left. If it points to the left, then the slope
  • #1
Poetria
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Homework Statement
Consider a block of mass 1 kg sitting on a plane inclined to an angle of theta = pi/6. Approximate the force due to gravity to be 10 N pointing straight down. Find the vector decomposition into tangent and perpendicular vector components by following the method above.
Relevant Equations
$$\vec v=\vec a+\vec b$$
$$\vec a = \vec g_{tangential}$$
$$\vec b = \vec g_{normal}$$
It's a puzzle. I have decomposed vector v by using formulas known from physics: m*g*sin(theta) and m*g*cos(theta).

I got: ##\vec v = (5, 5*\sqrt{3})##

But it has been marked as wrong. Consequently, the rest of my calculations is not correct. Could you tell me, why?
 
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  • #2
Are those components in the correct order?
 
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  • #3
I think its not a matter of correct order but of correct sign. Usually the positive direction is taken upwards, so I think at least one of the components should be negative.
 
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  • #4
Poetria said:
Approximate the force due to gravity to be N pointing straight down
If you have the question correct,then I would expect 'N' to be part of the answer.
 
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  • #5
It should be 10N.
 
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  • #6
Further thoughts...

What is wrong withthe following statement?
"The length of this piece of string is 250."

What is wrong with the following statement?
##\vec v = (5, 5*\sqrt{3})##

Also, the instructions say "by following the method above" but we don't know what this method is. For example, the required method could be to draw a scale diagram and take measurements, in which case you have not followed the instructions. (But if the required method is to use the formulae you quote, then that's OK.)
 
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  • #7
I do know that. The problem is that in a similar problem ##\vec v## was given. On the other hand, why should the solution be dependent on a method?
Well, I have the feeling that I haven't got the reasoning in terms of ##\vec u##.

$$\vec v = \vec a + \vec b$$

##\vec a## is a component of ##\vec v## in the ##\vec u## direction.
##\vec b## is a component of ##\vec v## penpedicular to the ##\vec u## direction.

Given ##\vec u## and ##\vec v## find ##\vec a## and ##\vec b##

##\vec a## is in the same direction as ##\vec u##, therefore

##\vec a## = ##\lambda*\vec u##

##\lambda = \frac {(\vec u*\vec v)} {(\vec u*\vec u)}##

##\vec a = \frac {(\vec u*\vec v)} {(\vec u*\vec u)}*\vec u##
 
  • #8
Are your answers marked incorrect by your teacher? Or are you entering the answers into a (software) teaching package?

If the latter, is some particular format required, e.g. 5.00N, 8.66N ?

The solution should not depend on the method - I agree. However, if instructed to use a specific method, you are being asked to demonstrate your knowledge of the method; so using a different method (even if it gives the correct answer) means you have not answered the question properly.
 
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  • #9
Steve4Physics said:
Are your answers marked incorrect by your teacher? Or are you entering the answers into a (software) teaching package?

If the latter, is some particular format required, e.g. 5.00N, 8.66N ?

The solution should not depend on the method - I agree. However, if instructed to use a specific method, you are being asked to demonstrate your knowledge of the method; so using a different method (even if it gives the correct answer) means you have not answered the question properly.
It's a software. The solution should be a vector. "Find a vector ##\vec g## ." Well, I am not comfortable with this method. But I don't know what I am missing.
There is an example:
"Decompose the vector ##\vec v (1,2)## into components that point in the direction of ##\vec u = (1,1)## and normal to ##\vec u##."

This exercise is a warm-up for multivariable calculus.
 
  • #10
It's possible the package wants a particular format for your answer. Is there any guidance? You could try these for example:
(5, 5√3)N
(5N, 5√3N)
(5, 8.66)N
(5N, 8.66N)

It's also possible that the software has a bug, so that your correct answer is being marked incorrect.

Also, if required, here's a video which should help with your example problem:
 
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  • #11
Steve4Physics said:
It's possible the package wants a particular format for your answer. Is there any guidance? You could try these for example:
(5, 5√3)N
(5N, 5√3N)
(5, 8.66)N
(5N, 8.66N)

It's also possible that the software has a bug, so that your correct answer is being marked incorrect.

Also, if required, here's a video which should help with your example problem:

Great. :) I will watch it. :)
 
  • #12
Just wondering if your try ##(5,-5\sqrt{3})## what do you get?
 
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  • #13
##\vec a = (5,0)##
##\vec b= (0, -5*\sqrt{3})##

Yeah, I also thought that the sign could be wrong.
 
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  • #14
Poetria said:
##\vec a = (5,0)##
##\vec b= (0, -5*\sqrt{3})##

Yeah, I also thought that the sign could be wrong.
I have tried. It has been marked as wrong. Well, I will remember this problem for the rest of my life, I suppose.
 
  • #15
Signs and units are important, but by convention the ##x## component comes first. And in this case is larger than the ##y## component.

Edit: ignore this!
 
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  • #16
PeroK said:
Signs and units are important, but by convention the ##x## component comes first. And in this case is larger than the ##y## component.
What axis do you take as the x-axis? The angle of the incline is ##\pi/6## not ##\pi/3##.
 
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  • #17
Delta2 said:
What axis do you take as the x-axis? The angle of the incline is ##\pi/6## not ##\pi/3##.
Ah, I was thinking about ##\vec v## being the velocity down the slope.
 
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  • #18
Poetria said:
##\vec a = (5,0)##
##\vec b= (0, -5*\sqrt{3})##
Those vectors are horizontal and vertical, so can't be correct. Try this…

I'll omit units (N) for readability.

Weight is ##\vec W## = <0, -10>.
I think we have to find:
##\vec W_1## = the vector representing component of weight parallel to the slope;
##\vec W_2## = the vector representing component of weight normal to the slope.

We aren’t told if the slope is uphill to the right (+x direction) or to the left (-x direction) so guess it is to the right (and remember an incorrect guess leads to an incorrect sign on x-components).

The direction of the slope (the vector parallel to the slope pointing uphill) is then:
##\vec S = <cos(\frac {\pi}{6}), sin(\frac {\pi}{6})> = <\frac {√3}{2}, \frac 1 2>##

Note that ##||\vec S||## = 1 (because sin²+cos² = 1) i.e. ##\vec S## is a unit vector. This can simplify/shorten the working but I’ll show the working in full.

Using the standard projection formula, the projection of W onto S gives:
##\vec W_1 = \frac {<0,-10>•<\frac {√3}{2}, \frac 1 2>} {(\frac {√3}{2})² + (\frac 1 2 )² }<\frac {√3}{2}, \frac 1 2>##
##= -5<\frac {√3}{2}, \frac 1 2>##
##=<\frac {-5√3}{2}, \ -\frac 5 2>##

(A quick check shows ##||\vec W_1|| = 5## as we’d expect.)

##\vec W_2## is then ##\vec W – \vec W_1## which you can complete.
 
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  • #19
Poetria said:
Homework Statement:: Consider a block of mass 1 kg sitting on a plane inclined to an angle of theta = pi/6. Approximate the force due to gravity to be 10 N pointing straight down. Find the vector decomposition into tangent and perpendicular vector components by following the method above.
Relevant Equations:: $$\vec v=\vec a+\vec b$$
$$\vec a = \vec g_{tangential}$$
$$\vec b = \vec g_{normal}$$

It's a puzzle. I have decomposed vector v by using formulas known from physics: m*g*sin(theta) and m*g*cos(theta).

I got: ##\vec v = (5, 5*\sqrt{3})##

But it has been marked as wrong. Consequently, the rest of my calculations is not correct. Could you tell me, why?
Assuming the pair ##(5, 5\sqrt 3)## refers to horizontal and vertical components, shouldn't ##\vec v## just be ##(0,-10)~\rm N##? The vectors ##\vec a## and ##\vec b## will have different components, but they should sum to a vector that points downward. In linear algebra speak, what basis are you using to express the vectors?
 
  • #20
vela said:
Assuming the pair ##(5, 5\sqrt 3)## refers to horizontal and vertical components, shouldn't ##\vec v## just be ##(0,-10)~\rm N##? The vectors ##\vec a## and ##\vec b## will have different components, but they should sum to a vector that points downward. In linear algebra speak, what basis are you using to express the vectors?
We thought originally that the basis is the vector S of post #18 and the normal vector to S. With that basis ##\vec{v}## is indeed written as ##(5,5\sqrt 3)## (correction of this up to a sign for each component).
However after reading more carefully post #18, it seems that the problem wanted us to find the vectors W1 and W2, expressed in the original basis that is the basis where ##\vec{v}=(0,-10)##,
 
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  • #21
Many thanks, I got it eventually. :)
I have another exercise similar to this as homework and it is clearly stated that the basis is where F = [0,-10].
 
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Related to Vector decomposition - gravity

1. What is vector decomposition?

Vector decomposition is the process of breaking down a vector into its component parts. This allows us to analyze the effect of each component on the overall vector.

2. How does vector decomposition relate to gravity?

In the context of gravity, vector decomposition is used to analyze the force of gravity acting on an object. The force of gravity can be broken down into its horizontal and vertical components, allowing us to understand the direction and magnitude of the force.

3. What is the formula for vector decomposition in relation to gravity?

The formula for vector decomposition in relation to gravity is Fx = Fcosθ and Fy = Fsinθ, where F is the force of gravity, θ is the angle of the vector with respect to the horizontal axis, and Fx and Fy are the horizontal and vertical components of the force, respectively.

4. Why is vector decomposition important in studying gravity?

Vector decomposition is important in studying gravity because it allows us to analyze the effects of gravity in different directions. This is particularly useful in understanding the motion of objects in free fall or on inclined planes, where the force of gravity may act at an angle.

5. How is vector decomposition used in real-world applications?

Vector decomposition is used in various real-world applications, such as engineering and physics. For example, in structural engineering, vector decomposition is used to analyze the forces acting on a building or bridge, while in physics, it is used to understand the motion of objects under the influence of gravity.

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