Work to Lift an Object

  • B
  • Thread starter jonasrosa
  • Start date
  • #1
jonasrosa
6
2
TL;DR Summary
If W= F*d*cos(θ), is it just going to be 0?
So, from what I remember, W=F*D*cos (θ). If I'm lifting, θ=90° and so, the cos = 0. So is the work just 0? Why? I still moved the object through a distance, which is the usual non-mathematical definition of Work.
 

Answers and Replies

  • #2
berkeman
Mentor
64,121
15,322
Welcome to PF. :smile:

How are you defining ##\theta## ? Work is force multiplied by distance through the path of the applied force. When you are lifting something, what is the direction and path of the applied force?
 
  • Like
Likes topsquark and jonasrosa
  • #3
jonasrosa
6
2
Welcome to PF. :smile:

How are you defining ##\theta## ? Work is force multiplied by distance through the path of the applied force. When you are lifting something, what is the direction and path of the applied force?
Ok, so, from what I remember ##\theta##? was the angle between the force and the object (meaning the angle at which I'm moving rhe object). So, horizontal movement would be a 0° angle and vertical would be 90°. It's been quite a while since I've studied physics, so I'm very rusty, but I was studying work today and I remember seeing this formula and it got me confused
 
  • #4
berkeman
Mentor
64,121
15,322
Yeah, so it sounds like you're trying to apply a "formula" for a situation where the variables are not defined correctly. It's better to try to understand the vector definition of Work, instead of trying to apply simplified formulas without understanding where they come from.

Have you had any introduction to vectors yet in your learning?
 
Last edited:
  • Like
Likes russ_watters, topsquark and erobz
  • #5
erobz
Gold Member
1,478
659
$$ W = \int \vec{F} \cdot d\vec{s}$$
 
  • #6
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
20,004
10,647
Ok, so, from what I remember ##\theta##? was the angle between the force and the object (meaning the angle at which I'm moving rhe object). So, horizontal movement would be a 0° angle and vertical would be 90°. It's been quite a while since I've studied physics, so I'm very rusty, but I was studying work today and I remember seeing this formula and it got me confused
You’ve got the angles backwards. To lift an object you apply a vertical force, not a horizontal one. The angle between force and displacement when lifting straight up is therefore zero.
 
Last edited:
  • Like
Likes vanhees71 and russ_watters
  • #7
jonasrosa
6
2
Yeah, so it sounds like you're trying to apply a "formula" for a situation where the variables are not defined correctly. It's better to try to understand the vector definition of Work, instead of trying to apply simplified formulas without understanding where they come from.

Have you had any introduction to vectors yet in your learning?
I have, but it was over 5 years ago. I am trying to understand what the formula means and why it doesn't apply on this situation or what am I misinterpreting, because once I realized this, it felt very weird
 
  • #8
jonasrosa
6
2
You’ve got the angles backwards. To lift an object you apply a vertical force, not a horizontal one. The angle between force and displacement when lifting straight up is therefore zero.
Ok, I think I get it. So, what matters is the angle of the force in relation to the movement? So, if I'm applying vertical force to move something vertically, the angle is 0 and if I apply horizontal force to move something horizontally, the angle is also 0?
 
  • #9
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,710
15,325
Ok, I think I get it. So, what matters is the angle of the force in relation to the movement? So, if I'm applying vertical force to move something vertically, the angle is 0 and if I apply horizontal force to move something horizontally, the angle is also 0?
Yes.
 
  • #10
jonasrosa
6
2
Yes.
Ok, now it makes sense. Thanks a lot.
 
  • #11
russ_watters
Mentor
22,050
9,148
Ok, now it makes sense. Thanks a lot.
FYI, this would all be a lot clearer if you drew a diagram and labeled the force, direction of motion and angle.
 
  • #12
jonasrosa
6
2
FYI, this would all be a lot clearer if you drew a diagram and labeled the force, direction of motion and angle.
Yes, probably. Didn't think of doing that. Thanks a lot
 
  • #13
bob012345
Gold Member
1,829
793
Ok, I think I get it. So, what matters is the angle of the force in relation to the movement? So, if I'm applying vertical force to move something vertically, the angle is 0 and if I apply horizontal force to move something horizontally, the angle is also 0?
Also, realize that if the force is opposed to the motion such as a block is moving to the right but a force on it points to the left, the angle is 180° and the cosine is -1. That means the block is slowing down.
 
  • Like
Likes russ_watters

Suggested for: Work to Lift an Object

Replies
34
Views
1K
  • Last Post
Replies
2
Views
361
  • Last Post
Replies
1
Views
281
Replies
8
Views
828
Replies
3
Views
481
Replies
23
Views
615
Replies
34
Views
2K
  • Last Post
2
Replies
61
Views
3K
Replies
2
Views
977
Replies
5
Views
4K
Top