# Work to Lift an Object

• B
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.

Mentor
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?

topsquark and jonasrosa
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

Mentor
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.

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russ_watters, topsquark and erobz
Gold Member
$$W = \int \vec{F} \cdot d\vec{s}$$

berkeman
Staff Emeritus
Homework Helper
Gold Member
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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vanhees71 and russ_watters
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.

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?

berkeman
Homework Helper
Gold Member
2022 Award
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.

berkeman
Mentor
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.

berkeman
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

Gold Member
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.

russ_watters