# Forces and inclined planes

1. Jun 28, 2014

### jonroberts74

Suppose a force F is acting downwards on an object sitting on a plane that is inclined 45 degrees to horizontal. express the force as a sum of a force acting parallel to the plane and on acting perpendicular to the plane

I figured I need to use ||F||cos 45i+ ||F||sin 45j

and the angle between the plane and x-axis will the be the same angle between the downward and perpendicular

this is where I am getting stuck

2. Jun 28, 2014

### LCKurtz

You need a picture and a coordinate system. For example if your force is pointing straight down you would have $\vec F = -\|\vec F\|j$. In your drawing what do you get for the vectors parallel and perpendicular to the plane. You need those next.

3. Jun 28, 2014

### jonroberts74

if [TEX]\vec{F} = -||F||j[/TEX]

then [TEX]\vec{F}[/TEX] dotted with [TEX]<1,0>[/TEX] would be zero

the angle between <1,0> and the vector parallel to the plane wouldn't that be [TEX]cos \theta = \frac{\vec{a}*<1,0>}{||a|| ||<1,0>||} [/TEX] where vector a is the parallel to the plane

I don't know if I am getting closer

sorry, it the wrong slash key, I am entering it in properly but it's not working

Last edited: Jun 28, 2014
4. Jun 28, 2014

### LCKurtz

5. Jun 28, 2014

### jonroberts74

I used the correct brackets and it still is not working

6. Jun 29, 2014

### LCKurtz

I fixed your tex in my quote above.

Anyway the vector <1,0> doesn't have anything to do with this. What I asked you to do in my first reply was to give two vectors, one down the plane and one perpendicular to it. You should be able to do that from your picture. Don't give me unknown vectors $a$ or sines and cosines. Give me explicit vectors with numbers. While you are at it make them unit vectors. The rest will be easy after that.

7. Jun 29, 2014

### jonroberts74

<1,1> would be one parallel to the plane and <1,-1> would be perpendicular to the plane

$$\vec{F} = <1,1> + <1,-1>$$

??

8. Jun 29, 2014

### LCKurtz

No, that isn't how you resolve F.

I think you want <-1,-1> parallel but headed downward. Make them unit vectors and call the unit vectors $p$ and $n$ for parallel and normal.

Now remember that the scalar component of a vector $a$ in the direction of a unit vector $\hat d$ is given by $a\cdot \hat d$ and the vector projection in that direction is $(a\cdot \hat d)\hat d$. Use that formula to get your resolution of F.

I will be gone from this thread until tomorrow sometime.

9. Jun 29, 2014

### jonroberts74

I don't know anything about resolving forces.

$$<1,1>-<1,-1>=<0,2>$$

would be be a sum of the parallel and perpendicular resulting in the downward force

I know the projection formula $$\vec{p} = \frac{\vec{a}*\vec{v}}{||a||^2}\vec{a}$$

$$\vec{p} = <\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}> ; \vec{n} = <\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}>$$

I didn't do <-1,-1> because that seems perpendicular to the plane and not parallel

Last edited: Jun 29, 2014
10. Jun 29, 2014

### CAF123

See the attached figure. In a coordinate system with vertical (y) and horizontal (x) directions then the vector F can be described by $\vec F = F \hat{y}$ since it is pointing down and there is no horizontal component.

Your job is to rewrite the same vector F in a new coordinate system defined with axes parallel and perpendicular to the plane, denoted by the bold lines in the sketch. Convince yourself that angle x is also 45 degrees. Then, to find the projections of F onto the directions parallel and perpendicular to the plane, it boils down to finding the 'adjacent' and 'opposite' sides of the triangle I have shown. Can you find these?

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11. Jun 29, 2014

### LCKurtz

That is the same as the formula I gave where $\frac {\vec a}{\|\vec a\|} = \hat a$

If <1,1> is parallel to the plane then so is <-1,-1>. You want whichever is in the right direction to fit your picture. You need to use your projection formula on the parallel and perpendicular vectors to get the two forces you want. Your $\vec v$ is your given force vector and the $a$ in your formula is either your $\vec p$ or $\vec n$ vectors.

12. Jun 29, 2014

### jonroberts74

<-1,-1> would be the given for the parallel

I did the projection for that and the parallel (I had it all typed out but couldn't get the latex to work on this website, never have trouble with it anywhere else)

getting

$$<\frac{-1}{2}, \frac{-1}{2}>$$

this seems to be getting overly complicated.

or is v the initial downward force which would be -j

13. Jun 29, 2014

### jonroberts74

I checked the book, it gives the answer as

$$\vec{F} = (||F||v+||F||h)/\sqrt{2}$$

where $$v=(i-j)/\sqrt{2}$$ and $$h=-(i+j)/\sqrt{2}$$

14. Jun 29, 2014

### vela

Staff Emeritus
I've attached a diagram, which is something I hope you have drawn for yourself. The normal unit vector $\hat{n}$ points in the opposite direction than yours. (I was too lazy to fix the picture, but it doesn't make a difference anyway.) I'm assuming the positive direction for the x' axis is up and to the right, and for the y' axis, up and to the left.

$\vec{A}$ and $\vec{B}$ are the components of $\vec{F}$ along the rotated axes. You can see that both components point in the negative direction on their respective axes.

According to the projection formula, you have
$$\vec{A} = (\vec{F}\cdot\hat{p})\hat{p} = \left(\langle 0,-F \rangle \cdot \left\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle\right) \hat{p}.$$ What do you get when you work that out?

Do the same thing using $\hat{n}$ to find the normal component, $\vec{B}$.

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15. Jun 29, 2014

### jonroberts74

so

$$<0,-F/\sqrt{2}>\hat{p}$$

$$<0, \frac{-F}{2\sqrt{2}}>$$

??

that doesn't seem related to the answer the book gives

16. Jun 29, 2014

### vela

Staff Emeritus
No. You need to look up how to calculate a dot product.

17. Jun 29, 2014

### jonroberts74

$$<\frac{-F}{2},\frac{-F}{2}>$$

$$\hat{p} = <\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}>$$ this is what I had for $$\hat{p}$$ before

I know how to do the dot product, I had a lapse in what was going on there. this doesn't seem any closer to the answer

18. Jun 29, 2014

### vela

Staff Emeritus
What is this supposed to be? $\vec{A}$?

19. Jun 29, 2014

### LCKurtz

Once you put the minus signs in your $\vec p$, don't the $v$ and $h$ in the book's answer look a lot like they have used your $\vec n$ and $\vec p$ in the calculation? You are getting close.

Last edited: Jun 29, 2014
20. Jun 29, 2014

### jonroberts74

$$<\frac{-1||F||}{\sqrt{2}}, \frac{-1||F||}{\sqrt{2}}> + <\frac{||F||}{\sqrt{2}, \frac{-1||F||}{\sqrt{2}}>$$

I can't get this to come up properly, it has proper tags, not sure why it won't show up properly