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?

 

[PeroK]

Are those components in the correct order?

 

[Delta2]

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.

 

[Steve4Physics]

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.

 

[Poetria]

It should be 10N.

 

[Steve4Physics]

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]

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

$\vec a$ = $\lambda*\vec u$

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

$\vec a = \frac {(\vec u*\vec v)} {(\vec u*\vec u)}*\vec u$

 

[Steve4Physics]

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.

 

[Poetria]

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.

 

[Steve4Physics]

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]

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. :)

 

[Delta2]

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

 

[Poetria]

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

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

 

[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.

 

[PeroK]

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!

 

[Delta2]

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]

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.

 

[Steve4Physics]

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

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.

 

[vela]

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?

 

[Delta2]

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]

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].