Maximum Distance for Man to Stretch Spring in Equilibrium

In summary: Fx = 0 is not satisfied because fs is not equal to Tcosθ. The friction force is static, not kinetic. This means that the man is not slipping. The friction force is static friction. This means that the force of friction is equal and opposite to the direction of the force trying to make the man slip. The friction force is necessary for the man to walk to the right. Without friction, he would not be able to move to the right. In summary, the man with mass M attached to a spring with constant k and angle θ can be moved slowly until the normal reaction force vanishes. The maximum distance he can stretch the spring beyond its natural length is given by
  • #1
decentfellow
130
1

Homework Statement


A man with mass ##M## has its string attached to one end of the spring which can move without friction along a horizontal overhead fixed rod. The other end of the spring is fixed to a wall. The spring constant is ##k##. The string is massless and inextensible and mantains a constant angle ##\theta## with the overhead rod, even when the man moves. There is friction with coefficient ##\mu## between the man and the ground. What is the maximum distance (in ##\text{m}##) that the man moving slowly can stretch the spring beyond its natural length.
scan0003.jpg


Homework Equations


I can't seem to figure out what to write in this so as usual I will be writing this ##\vec{F}=m\vec{a}##

The Attempt at a Solution


Now as the string is constrained to move while making an angle of ##\theta## with the overhead rod so we get the relation
$$T\cos\theta=kx$$

geogebra-export.png

geogebra-export (1).png

As the man moves slowly it means that he is always in equilibrium. Therefore the man moves till the normal reaction vanishes, i.e. till ##N=0##, because if he moves any further then he would have exert a force on the string this will not let the rope maintain its angle ##\theta## with the overhead rod.

$$T\sin\theta=Mg\implies x=\dfrac{Mg}{k\tan\theta}$$

But on thinking on the answer provided by the book, which is ##x=\dfrac{\mu Mg}{k(1+\mu\tan\theta)}##, I came to conclude that what the book assumes is that that the friction is acting in the opposite direction then that what I had assumed which can be seen in the last figure above. My reasoning, for that direction was that as the man moves rightward there is relative motion with the ground, so the friction resists this(I kinda assumed his slowly walking to be sliding, or how else would he remain in equilibrium).

Also, if the man does move further rightwards from the book's critical point of equilibrium, as the man is always to remain in equilibrium, so he would just generate so much force to be in equilibrium and not let the spring force pull him backwards as he moves rightwards.

So, according to me the man can move only till the normal reaction force vanishes.

Please correct me at the places where my assumptions are wrong.
 
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  • #2
decentfellow said:
View attachment 105715
As the man moves slowly it means that he is always in equilibrium.
Yes, he is in equilibrium. So, ΣFx = 0 and ΣFy = 0 for the man. If the friction force is in the direction shown in your figure, then you can see that ΣFx = 0 is not satisfied.

I kinda assumed his slowly walking to be sliding
I think this is where you are having trouble. The horizontal component of the force T acting on the man tries to slide the man to the left. The static friction prevents the man from slipping to the left.

So, according to me the man can move only till the normal reaction force vanishes.
If the normal force were to go to zero, what would happen to the friction force? Could the man be in equilibrium if the normal force approaches zero?
 
  • #3
The man can move if the friction must be static friction.
We have : $$N=Tsin\theta -mg$$
##F_{fr}=Tcos\theta≤ \mu N =\mu(Tsin\theta -mg)##
So we have ##x=\dfrac{\mu Mg}{k(1+\mu tan\theta)}##
 
  • #4
  • #5
TSny said:
Yes, he is in equilibrium. So, ΣFx = 0 and ΣFy = 0 for the man. If the friction force is in the direction shown in your figure, then you can see that ΣFx = 0 is not satisfied.
But what I have read about slowly moving something is that that it is moved with same net force plus some little extra(which tends to 0) that acts on it just in the other direction of the net force so that it moved with approxiamtely zero force, like in the case of deriving the potential of an electric charge when we derive the potential of a point we move the charge from infinity to the concerned point slowly. So, when the man moves slowly he has some relative motion with the ground, so doesn't that mean that he experiences friction in the direction opposite to where he moves, because either the man moves due to the force that he exerts by himself(assuming that he moves by sliding) or some external agency exerts on him to move him. I think the point that I do not get here is which force makes the man move, if I get that then I think I can get where the friction force acts on the man.

TSny said:
If the normal force were to go to zero, what would happen to the friction force? Could the man be in equilibrium if the normal force approaches zero?
That's why he can't move further than the point where the normal reaction becomes 0.
 
  • #6
Hamal_Arietis said:
View attachment 105710
T is wrong. Must be reverse direction.
Im not sure my English is right.
Oops. That was a fault on my side, but let it leave it as it is cause it will be a rather lengthy process to redirection the arrow.
 
  • #7
If the man were to somehow slide to the right, then you would be correct that there would be a kinetic friction force to the left. However, the man is not sliding to the right. There is no slipping of his feet (until he tries to surpass the maximum distance he can attain). The friction force is static friction.

Consider the simplified picture below where the tension T in the spring acts horizontally on the man.
upload_2016-9-10_11-44-6.png


To walk to the right, the man must push his feet down and to the left on the floor. The reaction force of the floor on the man is up and to the right. The upward part of the force of the floor on the man is the normal force N. The rightward part of the force of the floor on the man is the static friction force fs. So, the forces acting on the man are as shown in the force diagram above.

You can see that the normal force would never go to zero here.
 
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  • #8
TSny said:
If the man were to somehow slide to the right, then you would be correct that there would be a kinetic friction force to the left. However, the man is not sliding to the right. There is no slipping of his feet (until he tries to surpass the maximum distance he can attain). The friction force is static friction.

Consider the simplified picture below where the tension T in the spring acts horizontally on the man.
View attachment 105718

To walk to the right, the man must push his feet down and to the left on the floor. The reaction force of the floor on the man is up and to the right. The upward part of the force of the floor on the man is the normal force N. The rightward part of the force of the floor on the man is the static friction force fs. So, the forces acting on the man are as shown in the force diagram above.

You can see that the normal force would never go to zero here.
Okay so what I have got from what you said and what I read here https://www.physicsforums.com/threads/confused-on-static-friction-walking-and-driving.240422/ is that the reaction force which pushes the man forward while walking is none other than the static friction and I had been wrongly interpreting the walking part as sliding, so now what I concluded is that the man can slowly "walk" forward only when he has the sufficient net force propelling him forward which comes from ##f_s-T\cos\theta##. Now just for the sake of satiating my curiosity what would have been the distance that he would have been able to traverse had he been sliding would it have been the point where the normal reaction vanishes.
 
  • #9
decentfellow said:
Okay so what I have got from what you said and what I read here https://www.physicsforums.com/threads/confused-on-static-friction-walking-and-driving.240422/ is that the reaction force which pushes the man forward while walking is none other than the static friction and I had been wrongly interpreting the walking part as sliding, so now what I concluded is that the man can slowly "walk" forward only when he has the sufficient net force propelling him forward which comes from ##f_s-T\cos\theta##.
Yes. Since he is assumed to walk slowly without any acceleration, the net force propelling him can be taken to equal zero.

Now just for the sake of satiating my curiosity what would have been the distance that he would have been able to traverse had he been sliding would it have been the point where the normal reaction vanishes.
If he is sliding to the right, then something or someone must have given him an initial velocity to the right. The distance he would slide to the right would depend on the initial velocity (as well as on k, μk, M, and g). If you are considering the initial problem where T is at the angle θ, the normal force would go to zero only if the initial velocity was greater than or equal to a certain critical value.
 
  • #10
TSny said:
If he is sliding to the right, then something or someone must have given him an initial velocity to the right. The distance he would slide to the right would depend on the initial velocity (as well as on k, μk, M, and g). If you are considering the initial problem where T is at the angle θ, the normal force would go to zero only if the initial velocity was greater than or equal to a certain critical value.
Alright that was just the thing that I wanted, thanks for all the time and the hardwork you put into make me understand:wink:
 

Related to Maximum Distance for Man to Stretch Spring in Equilibrium

1. What is the maximum distance a man can walk?

The maximum distance a man can walk varies depending on individual factors such as age, fitness level, and terrain. However, on average, a healthy adult male can walk up to 20-30 miles in a day.

2. Can a man walk to the moon?

No, it is currently not possible for a man to walk to the moon. The distance between Earth and the moon is approximately 238,855 miles, which would take over 19 years of non-stop walking at an average pace to cover.

3. How far can a man walk without stopping?

The distance a man can walk without stopping also varies depending on individual factors. However, an average adult can walk continuously for up to 6-8 hours before needing to take a break for rest and hydration.

4. What is the farthest distance a man has ever walked?

The farthest distance a man has ever walked is approximately 15,000 miles, achieved by George Meegan in 1977. Meegan walked from the southern tip of South America to the northernmost point in Alaska over the course of 2,425 days.

5. Can a man walk endlessly without getting tired?

No, humans are not capable of walking endlessly without getting tired. Walking requires energy and physical exertion, and the body needs rest and nourishment to sustain it. Eventually, fatigue and exhaustion will set in, and the body will need to rest.

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