# Vector decomposition - gravity

Poetria
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?

Last edited:

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Are those components in the correct order?

• Poetria
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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.

• Poetria
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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.

• • vela, Delta2 and Poetria
Poetria
It should be 10N.

• Delta2
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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.)

• Poetria
Poetria
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##

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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.

• Poetria
Poetria

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.

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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:

• Poetria
Poetria
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. :)

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Just wondering if your try ##(5,-5\sqrt{3})## what do you get?

• Poetria
Poetria
##\vec a = (5,0)##
##\vec b= (0, -5*\sqrt{3})##

Yeah, I also thought that the sign could be wrong.

• Delta2
Poetria
##\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.

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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!

Last edited:
• Poetria
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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##.

• PeroK
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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.

• Delta2
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##\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.

• • Delta2, docnet and 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?
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?

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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)##,

• Poetria and Steve4Physics
Poetria
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].

• Delta2