Need help with Newton's Laws, basic Physics

AI Thread Summary
The discussion focuses on clarifying concepts related to Newton's Laws and the definition of work in physics. It explains that using wheels or rollers reduces friction because rolling does not involve sliding, and the resistance encountered is mainly due to deformation rather than friction. The definition of work, expressed as Work = Force * distance, is highlighted as useful but not necessarily intuitive, emphasizing that work requires energy to move an object. Additionally, when pushing an object, maintaining a constant force will lead to acceleration, but the average force needed for work done is determined by the total distance and work, regardless of acceleration. Overall, the conversation aims to demystify these fundamental physics concepts for better understanding.
stfz
Messages
35
Reaction score
0
Hi,
I am taking a high school physics course, and there are a few points at which I am getting a bit of confusion.

My first area of confusion relates to Newton's Laws. There are a few scenarios in which I am not clear as to the interaction of forces, etc.

a) Exactly why does putting wheels/rollers/etc under a heavy object make it easier to move?
since force due to friction is : f = u*Fn, where u is the coefficient of friction, Fn is weight.
Looks like force due to friction is dependent only on weight and the coefficient of friction, not surface area which is in contact. What exactly makes it easier? Or is coefficient of friction dependent on surface area?

b) Is the definition of Work (Work (Joules) = Force * distance, Joules = Newton * m) supposed to be intuitive? Since I don't really understand how "work" is defined in that way. I can work with the definition, but it's just that I cannot REALLY understand why it is so.

Definition of work is : W = F * Δx, F = force, Δx = Distance. I just want to clarify this :
Assuming a zero-friction environment, if I push an object with constant force, the object will continually accelerate.
In order to produce work with an object for n meters, then, I assume I will need to push the object continually with constant force. This means I will have to keep going faster and faster. Is that right?

My textbook has this question : "A refrigerator repairman pushes a refrigerator 75 cm horizontally to get access to the panel behind it. If the worker pushed horizontally and did 175 Joules of work, with what force did he push?".

I assume that in this question, ignoring friction, then, that the repairman would have to push the fridge with constant force, thus giving it increasing velocity?

I'm sorry if this seems overly basic... my book did not explain this very concisely.
Would appreciate any input :)
Thanks!

Stephen
 
Physics news on Phys.org
(a) I think that rolling is quite a difficult idea. One essential point is that no work is done against friction, because there is no sliding.

(b) I think you've asked a very good question about the definition of work. I don't think it is intuitive. A teacher or textbook needs to provide a motivation for the definition. Mine would be that you have to 'pay' when work is done, whereas you don't if the force isn't moving. For example, if you wanted to keep a heavy object 2 metres above the floor, you could keep it on a platform (for hundreds of years if you wanted) with no expenditure on any sort of fuel. But if you wanted to lift the object by a further distance, you couldn't do this without some sort of fuel (such as the chemical energy used when muscles operate).

(c) Although you don't need to go into this in order to answer the question, I would imagine that the force exerted by the repairman is simply an equal and opposite force to the resistive force on the fridge. There will be a very small extra force needed to accelerate the refrigerator from rest to the very low speed at which he will slide it. When you calculate the work done, you take the force specified, that is the force exerted by the repairman, NOT the resultant force on the refrigerator (which is almost zero).
 
  • Like
Likes 1 person
stfz said:
a) Exactly why does putting wheels/rollers/etc under a heavy object make it easier to move?
since force due to friction is : f = u*Fn, where u is the coefficient of friction, Fn is weight.
Looks like force due to friction is dependent only on weight and the coefficient of friction, not surface area which is in contact. What exactly makes it easier? Or is coefficient of friction dependent on surface area?
Friction coefficient is irrelevant in rolling, because there is no sliding. In rolling the resistance comes from deformation of wheel & surface, which causes the wheel to be rolling slightly "uphill" all the time. It’s not always easier than sliding. The rolling resistance can be higher than sliding friction in some cases.

stfz said:
Is the definition of Work (Work (Joules) = Force * distance, Joules = Newton * m) supposed to be intuitive?
No, it supposed to be useful and consistent with other definitions of Energy.

stfz said:
Assuming a zero-friction environment, if I push an object with constant force, the object will continually accelerate.
In order to produce work with an object for n meters, then, I assume I will need to push the object continually with constant force. This means I will have to keep going faster and faster. Is that right?
Yes

stfz said:
My textbook has this question : "A refrigerator repairman pushes a refrigerator 75 cm horizontally to get access to the panel behind it. If the worker pushed horizontally and did 175 Joules of work, with what force did he push?". I assume that in this question, ignoring friction, then, that the repairman would have to push the fridge with constant force, thus giving it increasing velocity?
It doesn't matter if it accelerates, or moves at constant speed with friction. The average force depends only on the distance and the work done.
 
  • Like
Likes 1 person
Thanks! That helped
 
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