Doubt in the basics definition of work

AI Thread Summary
The discussion centers on the definition of work in physics, specifically the relationship between force and displacement. It clarifies that work is calculated using the dot product of the force vector and the displacement vector, which captures only the component of force that acts in the direction of the displacement. An example illustrates that if a force is applied perpendicular to the direction of movement, no work is done in that direction. The conversation also touches on the geometric interpretation of the dot product, emphasizing its role in determining the effective force contributing to work. Understanding this concept is crucial for accurately applying the principles of work in physics.
ehabmozart
Messages
212
Reaction score
0
Hello!

In my studying of work I've always been told that work done (by what? ) = Force (component along its path) times the distance it moves. Mathematically, W=F.s ... My question here is what is the geometrical presentation of saying Force x Distance ... In other words, when we multiply force vector times the displacement vector, what does it mean. If for example dot product the force vector by the unit vector of say the x-axis < 1, 0 ,0 > we will get the x component of the force. What is the case for work??

Thanks a lot for bearing my lengthy piece. And thanks in advance to whoever gives me a kind hand
 
Physics news on Phys.org
ehabmozart said:
If for example dot product the force vector by the unit vector of say the x-axis < 1, 0 ,0 > we will get the x component of the force. What is the case for work??
Take the dot product of the force and the displacement vectors. You are essentially taking the component of the force parallel to the displacement and multiplying it by the displacement.
 
The idea is that, if an object is moving to the east, and you apply a force on the object directed to the north, the object continues moving to the east at the same speed; so no work is done by the force as far as the movement to the east is concerned. But, the object is accelerating to the north, so the force is doing work on the object in that direction. So the conclusion is that only the component of movement parallel to the direction of the force results in work being done.

Chet
 
ehabmozart said:
My question here is what is the geometrical presentation of saying Force x Distance
That would be the cross product. But work is the dot product.

ehabmozart said:
If for example dot product the force vector by the unit vector of say the x-axis < 1, 0 ,0 > we will get the x component of the force.
Yes, that is the geometrical presentation of a dot product.
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top