Lifting an Object: Work, Force, and Cosine

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Discussion Overview

The discussion revolves around the concept of work in physics, specifically the relationship between force, distance, and the angle of application when lifting an object. Participants explore the implications of the formula W=F*D*cos(θ) in various contexts, including vertical and horizontal movements.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions whether work is zero when lifting an object, noting that the angle θ is 90° and thus cos(90°) = 0.
  • Another participant emphasizes the importance of defining the angle θ correctly, suggesting that it should represent the angle between the force applied and the direction of movement.
  • Some participants argue that the angle of the applied force is crucial, stating that when lifting vertically, the angle is actually 0° because the force and displacement are in the same direction.
  • There is a suggestion that understanding the vector definition of work may clarify the confusion surrounding the formula.
  • A later reply points out that if the force opposes the motion, such as a block moving right while a force acts left, the angle would be 180°, resulting in negative work.

Areas of Agreement / Disagreement

Participants express differing views on the correct application of the work formula and the definition of the angle θ. While some clarify that the angle should be 0° when lifting vertically, others initially suggest that the angle is 90°. The discussion remains unresolved regarding the initial confusion about the formula's application.

Contextual Notes

Participants mention a lack of recent experience with vectors and physics concepts, which may contribute to misunderstandings. There is also a suggestion that visual aids, such as diagrams, could enhance understanding.

Who May Find This Useful

This discussion may be useful for individuals studying physics concepts related to work, force, and motion, particularly those seeking clarification on the application of formulas in different contexts.

jonasrosa
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TL;DR
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.
 
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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?
 
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berkeman said:
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
 
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?
 
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$$ W = \int \vec{F} \cdot d\vec{s}$$
 
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jonasrosa said:
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.
 
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berkeman said:
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
 
Orodruin said:
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?
 
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jonasrosa said:
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.
 
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PeroK said:
Yes.
Ok, now it makes sense. Thanks a lot.
 
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  • #11
jonasrosa said:
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.
 
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  • #12
russ_watters said:
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
jonasrosa said:
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.
 
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