Lifting an Object: Work, Force, and Cosine

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In summary: 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.
  • #1
jonasrosa
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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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  • #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?
 
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  • #3
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
 
  • #4
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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  • #5
$$ W = \int \vec{F} \cdot d\vec{s}$$
 
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  • #6
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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  • #7
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
 
  • #8
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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  • #9
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.
 
  • #10
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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Related to Lifting an Object: Work, Force, and Cosine

1. How is work defined in the context of lifting an object?

Work is defined as the amount of force applied to an object multiplied by the distance the object is moved in the direction of the applied force.

2. What is the relationship between work and force when lifting an object?

The relationship between work and force when lifting an object is that the more force that is applied to an object, the more work is done. This means that the greater the force, the greater the amount of work required to lift the object.

3. How does the angle of the applied force affect the amount of work done when lifting an object?

The angle of the applied force affects the amount of work done when lifting an object because only the component of the force that is in the direction of the object's motion will contribute to the work. This means that the greater the angle between the applied force and the direction of motion, the less work will be done.

4. What is the role of cosine in the calculation of work when lifting an object?

Cosine plays a crucial role in the calculation of work when lifting an object because it represents the ratio between the applied force and the direction of motion. This means that the cosine of the angle between the applied force and the direction of motion is used to determine the amount of force that is actually contributing to the work.

5. How can the concept of work be applied to lifting an object in real-life situations?

The concept of work can be applied to lifting an object in real-life situations by understanding that the amount of work required to lift an object is directly proportional to the weight of the object and the distance it needs to be lifted. This means that by increasing the force applied or decreasing the angle of the force, less work will be required to lift the object. This can be useful in tasks such as weightlifting, construction, and moving heavy objects.

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