B Lifting an Object: Work, Force, and Cosine

[jonasrosa]

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.

 

[berkeman]

Welcome to PF.

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?

 

[jonasrosa]

Welcome to PF.

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

 

[berkeman]

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?

 

[erobz]

$$ W = \int \vec{F} \cdot d\vec{s}$$

 

[Orodruin]

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.

 

[jonasrosa]

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

 

[jonasrosa]

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?

 

[PeroK]

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.

 

[jonasrosa]

Yes.

Ok, now it makes sense. Thanks a lot.

 

[russ_watters]

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.

 

[jonasrosa]

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

 

[bob012345]

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.