# Work on a plane with angles forces

• Zsmitty3
In summary, the problem asks how much work is done by F3 and F2 as they move .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.
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

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.

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?

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)?

Here's the diagram

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.

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

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.

## 1. What is the concept of "work" in relation to a plane with angled forces?

The concept of work in physics refers to the amount of energy needed to move an object in a specific direction. In the case of a plane with angled forces, work is the force applied to the plane multiplied by the distance it moves in the direction of the force.

## 2. How do you calculate the work done on a plane with angled forces?

To calculate the work done on a plane with angled forces, you need to use the formula W = Fdcosθ, where W is the work, F is the force applied, d is the distance moved in the direction of the force, and θ is the angle between the force and the displacement vector.

## 3. What are the units of work and how are they measured?

The units of work are Joules (J) in the International System of Units (SI). It can also be measured in other units such as kilogram meters squared per second squared (kg*m^2/s^2) or Newton meters (Nm).

## 4. Can work be negative on a plane with angled forces?

Yes, work can be negative on a plane with angled forces. This occurs when the force applied is in the opposite direction of the displacement, resulting in a negative value for work. This is known as negative work and represents energy being taken away from the system.

## 5. How does the angle between the force and displacement affect the work done on a plane?

The angle between the force and displacement affects the work done on a plane by changing the direction of the force. If the force and displacement are in the same direction, the angle is 0 and the work done is at its maximum. As the angle increases, the work done decreases until the angle reaches 90 degrees and the work done becomes 0.

• Introductory Physics Homework Help
Replies
17
Views
2K
• Introductory Physics Homework Help
Replies
28
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
960
• Introductory Physics Homework Help
Replies
2
Views
827
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
19
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
22
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
2K