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Homework Help: Till where can the man walk

  1. Sep 10, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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}##

    3. 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

    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.
  2. jcsd
  3. Sep 10, 2016 #2


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    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 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.

    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?
  4. Sep 10, 2016 #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)}##
  5. Sep 10, 2016 #4
  6. Sep 10, 2016 #5
    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.

    That's why he can't move further than the point where the normal reaction becomes 0.
  7. Sep 10, 2016 #6
    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.
  8. Sep 10, 2016 #7


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    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.

    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.
  9. Sep 10, 2016 #8
    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.
  10. Sep 10, 2016 #9


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    Yes. Since he is assumed to walk slowly without any acceleration, the net force propelling him can be taken to equal zero.

    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.
  11. Sep 10, 2016 #10
    Alright that was just the thing that I wanted, thanks for all the time and the hardwork you put in to make me understand:wink:
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