Work on a plane with angles forces

[Zsmitty3]

1.So the problem is actually two parts. There's a block on an inclined 30degree plane and 3 different forces are acting on it.F1 is parallel to the plane is looks as though is the resultant force going up the plane in front of the block. F2 is acting at an angle 30degrees from the incline. So if you were pushing straight on the block with a force parallel to the plane, this force is acting 30 degrees above that, looking to make a right angle with Fg. F3 is perpendicular to the plane, looking to make a 90 degree with the normal force. The problems wants to know how much work is done by F3 and F2 as it moves .6m along the surface of the inclined plane in the upward direction. The m is 2kg and g=10. F1=20N F2=40N F3=10N


2. W=fd, W=mgcos*d



3. I think you would just use the equation above but somethings telling me I might be able to us a dot product just on the forces given. I'm not too sure though because F1 wouldn't make sense there. Also they're all in the same (x) direction so I don't think the dot product would work

 

[Simon Bridge]

Mechanical work done by force ##\vec{F}## over displacement ##\vec{r}## is given by ##W=\vec{F}\cdot\vec{r}## - well done. In your case it is probably easier to just find the component of the force in the direction of the displacement, and multiply that by the magnitude of the displacement. That way you don't need to memorize specific formulas.

 

[Zsmitty3]

Would this be right?

F2=40.0N acting at a 30 angle on the block that's sitting on a 30 slope.

Would it really just be 40N*.6m= 24J?

 

[Zsmitty3]

I've attached the image. View attachment Forces.docx There should also be a 30-degree right below the F2.The force of F2 in the y-directions would be... I don't know where that 30° would go? Would it end up being 40tan^-1 (30°)= Force in the y-direction?

And then for F3 in the y-direction it would be 10cos(30)?

 

[Zsmitty3]

Here's the diagram

 

[Zsmitty3]

I don't know if this is correct but I just made a triangle out of F2 and found the side going in the y direction by using 40tan30°=23.09N*.6m that it moves and got 13.86J for the first problem.

Same for problem 2 but use 10cos30°=8.66*.6m=5.2J but that's in the opposite direction of the movement, which is up, so I made it -5.2J

Again I have no clue if this works or not. We didn't go over dot product much.

 

[Zsmitty3]

I thought about this some more and think I was half right. Instead I found out if the mass moves. .6 in the y direction, .6tan30°= how much it moves in the x direction which is .3464 m

So now you can do dot product of

F2= 40Nx+23.09Ny
.3464mx+.6my Which = 27.72J

Same for F3= 5Nx+8.66Ny
.3464mx+.6my= 6.928J


Still not sure if this is correct

 

[Simon Bridge]

F2= 40Nx+23.09Ny

... (a) do you mean to use i and j unit vectors instead of x and y in that?
(b) ... but, according to post #1, |F2| = 40N ... what is |F1| according to what you just wrote.

The lynchpin is just to use rotated coordinates.
Place your coordinate axes so that the x-axis points in the direction of F1 in your diagram.

Thus
##\vec{F}_1 = F_1\vec{\imath}##
... etc. and the question should be much clearer too.