What is the formula for finding work in two dimensions?

In summary: In this case, the angle between the vectors will be the difference between these two angles, which is ##\theta = (\overline{F_{||}}-F_{\text{right}})|\Delta\vec s|##.
  • #1
Edel Crine
89
12
Homework Statement
A force F=Fx ı +Fy j with Fx =50N and Fy =12N is exerted on a particle as the particle moves along the x axis from x=1.0m to x=-5.0m.
(a) Determine the work done by the force on the particle.
(b) What is the angle be- tween the force and the particle’s displacement?
Relevant Equations
W=F*x
Since we don't know how far does it move in y-axis, I assumed that only x-component does work.
a) W=Fx*Δx = 50N*(-6m) = -300 J

b) Θ = arctan(12/50) = 13.5°

But, I'm not sure that did I do this in right way...h
 
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  • #2
Edel Crine said:
... as the particle moves along the x-axis from x=1.0m to x=-5.0m.
To clarify (in my mind,) is the particle constrained to "... move along the x- axis"?

because

"Since we don't know how far does it move in y-axis"
Well`, is the particle constrained or not?
 
  • #3
lewando said:
To clarify (in my mind,) is the particle constrained to "... move along the x- axis"?

because

"Since we don't know how far does it move in y-axis"
Well`, is the particle constrained or not?
Ummm it gave us only the displacement of x-axis, but we know the two components of the force, so I'm not sure whether does it move in y-direction as well, but I think so...?
 
  • #4
Edel Crine said:
Ummm it gave us only the displacement of x-axis, but we know the two components of the force, so I'm not sure whether does it move in y-direction as well, but I think so...?
I think you are right to suppose it is constrained by some other force(s), so the motion is only in the x axis.
But think about the angle again. Draw a diagram showing the directions of force and displacement vectors.
 
  • #5
Edel Crine said:
Homework Statement:: A force F=Fx ı +Fy j with Fx =50N and Fy =12N is exerted on a particle as the particle moves along the x-axis from x=1.0m to x=-5.0m.
(a) Determine the work done by the force on the particle.
(b) What is the angle be- tween the force and the particle’s displacement?
Relevant Equations:: W=F*x

Since we don't know how far does it move in y-axis, I assumed that only x-component does work.
a) W=Fx*Δx = 50N*(-6m) = -300 J

b) Θ = arctan(12/50) = 13.5°

But, I'm not sure that did I do this in right way...h
My reading of the problem statement is that the particle literally moves along the x-axis. The y-coordinate there is zero. So its displacement has a y-component of zero.

By the way, I see that you used very nearly the same title for two threads you posted today. That's a bad idea.
 
  • #6
SammyS said:
My reading of the problem statement is that the particle literally moves along the x-axis. The y-coordinate there is zero. So its displacement has a y-component of zero.

By the way, I see that you used the same title for two threads you posted today. That's a bad idea.
Oh yeah I did... 'Appreciate
 
  • #7
SammyS said:
My reading of the problem statement is that the particle literally moves along the x-axis. The y-coordinate there is zero. So its displacement has a y-component of zero.

By the way, I see that you used very nearly the same title for two threads you posted today. That's a bad idea.
Then, I think I'm good for part a), but I'm not sure for part b)...
 
  • #8
haruspex said:
I think you are right to suppose it is constrained by some other force(s), so the motion is only in the x axis.
But think about the angle again. Draw a diagram showing the directions of force and displacement vectors.
Ummm, would it be 0...?
 
  • #9
Edel Crine said:
Ummm, would it be 0...?
Are you guessing now?

Did you follow @haruspex's suggestion?
 
  • #10
SammyS said:
Are you guessing now?

Did you follow @haruspex's suggestion?
I just used the x-component of the force and the displacement...
Let me try this again...
 
  • #11
Ummm I got 166°...?
 
  • #12
Edel Crine said:
Ummm I got 166°...?
Yes.
 
  • #13
haruspex said:
Yes.
Now, I see! I appreciate your help again...!
 
  • #14
Edel Crine said:
Homework Statement:: A force F=Fx ı +Fy j with Fx =50N and Fy =12N is exerted on a particle as the particle moves along the x-axis from x=1.0m to x=-5.0m.
(a) Determine the work done by the force on the particle.
(b) What is the angle be- tween the force and the particle’s displacement?
Relevant Equations:: W=F*x

Since we don't know how far does it move in y-axis, I assumed that only x-component does work.
a) W=Fx*Δx = 50N*(-6m) = -300 J

b) Θ = arctan(12/50) = 13.5°

But, I'm not sure that did I do this in right way...h
 

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  • #15
Bilbo B said:
(Image of a solution)
Your method for finding the work done is not effectively different from that in post #1.
For the angle, you have made the same mistake as in post #1.
 
  • #16
Hello Edel!
From the way you wrote your response to a, I think that you have got the correct result just because the displacement was negative. I might be wrong, though.

For a constant force and a finite displacement with a constant direction, the work is defined as ##W=\vec F\cdot\Delta\vec s##, where ##\vec F## is the vector of the force for which you want to find the work, and ##\Delta\vec s## is the displacement vector.

We have, then, that ##W=|\vec F||\Delta\vec s|\cos\theta=(|\vec F|\cos\theta)|\Delta\vec s|=\overline{F_{||}}\times|\Delta\vec s|##.
##\theta## is the smallest angle you can make between the two vectors; If you draw two segments starting from the same point, then, starting from one vector, you can either rotate right or left to reach the other one.
##\overline{F_{||}}## represents the "signed" force in the direction of the displacement, i.e it can be positive or negative, depending ##\cos\theta##. (one can also leave force as it is and think of the "signed" component of the displacement in the direction of force, and it is more useful when the displacement's direction is variable while that of the force is constant.)
If the angle between your force vector and your displacement vector is not in ##(-\frac{\pi}{2},\,\frac{\pi}{2})##, then ##\overline{F_{||}}\leq0##. It is ##0## if ##\theta## takes either ##\pi/2## or ##-\pi/2##, because ##\cos\pm\frac{\pi}{2}=0##.

Let's go back to your problem (she has already solved it, don't delete my comment mods):
Capture.PNG
$$\Delta\vec s=\vec x_f-\vec x_i=(-5.0-1.0)\hat\imath\\
\vec F=50\hat\imath+12\hat\jmath$$
You can see from the values of ##\vec F## that it points right from the up direction, and from the values of ##\Delta\vec s## that it points directly to the left direction. This tells us that the smallest angle between the two forces is obtuse, hence ##\overline{F_{||}}## is negative.
To calculate the work, just plug in the values: ##W=-|F_{||}||\Delta\vec s|=-1\times50\times6.0=-30.\times10^1\mathrm J##
To find the angle, also use the work formula, and I think that this was the aim of question b.
##W=|\vec F||\Delta\vec s|\cos\theta=-30.\times10^1##, thus ##\cos\theta=\frac{-30.\times10^1}{|\vec F||\Delta\vec s|}\Leftrightarrow\theta=\arccos\left(\frac{-30.\times10^1}{|\vec F||\Delta\vec s|}\right)\approx166.5^\circ\approx \left(17.\times10^1\right)^\circ##.
Sorry for the long post. 🤷‍♂️
For other usages, another definition of the dot product ##\vec u\cdot\vec v## is ##\left(u_xv_x+u_yv_y+u_zv_z+...\right)##. The ##+...## is for higher dimensions, you stop at what you have.
 
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Related to What is the formula for finding work in two dimensions?

1. What is work in two dimensions?

Work in two dimensions refers to the calculation of work done on an object that is moving in a two-dimensional plane. This involves considering both the magnitude and direction of the force applied to the object.

2. How is work in two dimensions different from work in one dimension?

In one-dimensional work, the force and displacement are in the same direction, making the calculation of work simpler. In two-dimensional work, the force and displacement vectors may be at different angles, requiring the use of trigonometric functions to calculate the work done.

3. What is the formula for calculating work in two dimensions?

The formula for work in two dimensions is W = Fd cosθ, where W is work, F is the force applied, d is the displacement, and θ is the angle between the force and displacement vectors.

4. What is the unit of measurement for work in two dimensions?

The unit of measurement for work in two dimensions is joules (J). This is the same unit used for work in one dimension.

5. How is work in two dimensions used in real-world applications?

Work in two dimensions is used in various real-world applications, such as calculating the work done by a force on an object moving along a curved path, calculating the work done by a force on an object moving up or down a ramp, and calculating the work done by a force on an object moving in a circular motion.

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