Resolving Forces on an Inclined Plane: Finding the Components of a Force

[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

 

[LCKurtz]

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

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.

 

[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

 

[LCKurtz]

Please edit your post using [/tex] brackets to end your tex to make it readable and preview it before posting.

 

[jonroberts74]

I used the correct brackets and it still is not working

 

[LCKurtz]

if ##\vec{F} = -||F||j##, then ##\vec{F}## dotted with ##<1,0>## would be zero. The angle between ##<1,0>## and the vector parallel to the plane wouldn't that be $$cos \theta = \frac{\vec{a}*<1,0>}{||a|| ||<1,0>||} $$ 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

Please edit your post using [/tex] brackets to end your tex to make it readable and preview it before posting.

I used the correct brackets and it still is not working

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.

 

[jonroberts74]

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

[tex]\vec{F} = <1,1> + <1,-1>[/tex]

??

 

[LCKurtz]

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

[tex]\vec{F} = <1,1> + <1,-1>[/tex]

??

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.

 

[jonroberts74]

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.

I don't know anything about resolving forces.

[tex]<1,1>-<1,-1>=<0,2>[/tex]

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

I know the projection formula [tex] \vec{p} = \frac{\vec{a}*\vec{v}}{||a||^2}\vec{a}[/tex]

[tex] \vec{p} = <\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}> ; \vec{n} = <\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}>[/tex]

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

 

[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?

 

[LCKurtz]

I don't know anything about resolving forces.


I know the projection formula [tex] \vec{p} = \frac{\vec{a}*\vec{v}}{||a||^2}\vec{a}[/tex]

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

[tex] \vec{p} = <\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}> ; \vec{n} = <\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}>[/tex]

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

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.

 

[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

[tex]<\frac{-1}{2}, \frac{-1}{2}>[/tex]

this seems to be getting overly complicated.

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

 

[jonroberts74]

I checked the book, it gives the answer as

[tex]\vec{F} = (||F||v+||F||h)/\sqrt{2}[/tex]

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

 

[vela]

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}##.

 

[jonroberts74]

so

[tex]<0,-F/\sqrt{2}>\hat{p}[/tex]

[tex]<0, \frac{-F}{2\sqrt{2}}>[/tex]

??

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

 

[vela]

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

 

[jonroberts74]

[tex]<\frac{-F}{2},\frac{-F}{2}>[/tex]

[tex]\hat{p} = <\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}>[/tex] this is what I had for [tex]\hat{p}[/tex] 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

 

[vela]

[tex]<\frac{-F}{2},\frac{-F}{2}>[/tex]

What is this supposed to be? ##\vec{A}##?

 

[LCKurtz]

[tex] \vec{p} = <\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}> ; \vec{n} = <\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}>[/tex]

I checked the book, it gives the answer as

[tex]\vec{F} = (||F||v+||F||h)/\sqrt{2}[/tex]

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

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.

 

[jonroberts74]

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.

[tex]<\frac{-1||F||}{\sqrt{2}}, \frac{-1||F||}{\sqrt{2}}> + <\frac{||F||}{\sqrt{2}, \frac{-1||F||}{\sqrt{2}}>[/tex]

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

 

[jonroberts74]

What is this supposed to be? ##\vec{A}##?

yes

I did the dot then the product of that by p hat

 

[vela]

OK, so you have that the parallel component of ##\vec{F}## is
$$\vec{A} = -\frac F2 \hat{i} -\frac F2 \hat{j} = -\frac{F}{\sqrt{2}} \hat p.$$ Doesn't that match what's shown in the picture? Isn't it the same as what the book says is ##\frac{\|F\|}{\sqrt{2}}\hat{h}##?

 

[vela]

[tex]<\frac{-1||F||}{\sqrt{2}}, \frac{-1||F||}{\sqrt{2}}> + <\frac{||F||}{\sqrt{2}}, \frac{-1||F||}{\sqrt{2}}>[/tex]

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

You were missing a closing }.

 

[jonroberts74]

OK, so you have that the parallel component of ##\vec{F}## is
$$\vec{A} = -\frac F2 \hat{i} -\frac F2 \hat{j} = -\frac{F}{\sqrt{2}} \hat p.$$ Doesn't that match what's shown in the picture? Isn't it the same as what the book says is ##\frac{\|F\|}{\sqrt{2}}\hat{h}##?

ah I see, physics is not my strength

[tex] \frac{F}{\sqrt{2}}\hat{n} - \frac{F}{\sqrt{2}}\hat{p}[/tex]

this leads into another question I have then, its similar so I don't see a point in making a new thread for it

2) an object is moving in the direction i + j is being acted upon by the force vector 2i + j, express this force as the sum of a force in the direction of motion and a force perpendicular to the direction of motion.

so do the projection of f onto i+j and projection of f onto i-j

[tex]\vec{F} = <\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}> + < \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j}>[/tex]

 

[SammyS]

...

this leads into another question I have then, its similar so I don't see a point in making a new thread for it

2) an object is moving in the direction i + j is being acted upon by the force vector 2i + j, express this force as the sum of a force in the direction of motion and a force perpendicular to the direction of motion.

so do the projection of f onto i+j and projection of f onto i-j

[tex]\vec{F} = <\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}> + < \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j}>[/tex]

You should make a new thread!

It's in the rules for this Forum.

 

[LCKurtz]

Not only that, but you should also show your steps instead of just giving an answer and expecting us to work the whole thing ourselves to see if your answer is correct.