1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with friction

  1. Sep 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A man of mass ##80## kg, stands. on a horizontal weighing machine, of negligible mass, attached to a massless platform P that slides down a ##37^{\circ}## incline. The weighing machine reads ##72## kg. Man is always at rest w.r.t weighing machine.

    (a) The vertical acceleration of the man.
    (b) The coefficient of kinetic friction ##\mu## between the platform and man.

    2. Relevant equations

    &f_k\le \mu_{k}mg\qquad\qquad \text{where, $\mu_{k}\rightarrow$ coefficent of static friction} \\
    &f_s\le \mu_{s}mg\qquad\qquad \text{where, $\mu_{s}\rightarrow$ coefficent of static friction} \\

    3. The attempt at a solution

    The first part of the question was pretty easily solve but its the second part where I am having the problem.
    $$m_Mg-N_1=m_Ma_P\sin{37^{\circ}}\qquad\qquad\text{where, $m_M$ is the mass of man} \\
    \implies (80-72)g=80a_P\left(\dfrac{3}{5}\right)\implies a_P=\dfrac{5}{3}\text{m/s$^2$}$$
    man FBD.png

    So, from the above FBD of the man standing on the platform we see that

    From the above equation, I concluded that for the platform and the man to not move relatively to each other its not necessary that the friction acting b/w the man and the platform be the maximum static friction, so I don't think we can find the coefficient of "kinetic friction", even though if it had been possible we could have only found the coefficient of static friction. Now, if we were to find the coefficient of the static friction shouldn't the question have also mentioned that the force of friction acting b/w the surfaces to make them move together is the maximum friction.

    Also, the coefficient of friction that I found is
    $$(f_s)_{max}=m_Ma_P\cos\theta\implies \mu_sN_1=80\left(\dfrac{5}{3}\right)\left(\dfrac{4}{5}\right)\\
    \implies \mu_s(72g)=80\left(\dfrac{5}{3}\right)\left(\dfrac{4}{5}\right)\implies \mu_s=\dfrac{4}{27}$$

    So, the coefficient of friction that I got is ##\mu_s=\dfrac{4}{27}##, whereas the book got the coefficient of friction as ##\mu=\dfrac{13}{24}##. So, can you please point out where am I going wrong.
    Last edited: Sep 8, 2016
  2. jcsd
  3. Sep 7, 2016 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That is completely correct ... it is possible the question wants the coeff kinetic friction between the platform and the slope.
  4. Sep 7, 2016 #3
    If that is so then there is the question of why my and the books answer differ.
  5. Sep 8, 2016 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Not sure you read Simon's suggestion carefully. The question statement is wrong. It wants you to find the coefficient of kinetic friction between the platform and the slope. You can just assume the friction between platform and man is sufficient to prevent sliding.
    On this basis, I get the book answer.
  6. Sep 8, 2016 #5
    I am very sorry for the late reply, but yes it seems like I did not properly read Simon's post, and after solving it myself according to what Simon suggested I did get the answer but I solved it in a different manner than the book suggested and I am having a very hard time understanding what the book did, so I will be showing my attempt and the books attempt and the steps that I did not understand in the book's solution.

    My solution:-
    man FBD.png
    From the above FBD of the man, we get
    $$m_Mg-N_1=m_Ma_P\sin{37^{\circ}}\implies a_P=\dfrac{5}{3}\text{m/s$^2$}\\
    f_s=m_Ma_P\cos{37^{\circ}}\implies f_s=80\times\dfrac{5}{3}$$

    Now, as the platform P is massless so, from its FBD we get the following equations:-
    $$N_2=f_s\sin\theta+N_1\cos\theta=640 N\\
    \mu_k N_2+f_s\cos\theta=N_1\sin\theta\implies \mu_k=\dfrac{13}{24}$$

    Textbook's Solution:-


    $$(80-72)g=ma_y\implies a_y=1\text{m/s$^2$}\\
    a_y=a\sin{37^{\circ}}=1\text{m/s$^2$}\implies a=\dfrac{5}{3}\text{m/s$^2$}$$

    Now apply Newton's second law on man in direction of acceleration. Note that ##x-##component of acceleration is due to friction.
    $$mg\sin{37^{\circ}}-\mu mg\sin{37^{\circ}}=m\times(5/3)\\
    \implies 6-8\mu=5/3\implies 6-(5/3)=8\mu\implies \mu=\dfrac{13}{24}$$

    According to the FBD that the book has drawn is it not that that the book is finding the coefficient of friction b/w man and the platform not b/w platform and incline.
  7. Sep 9, 2016 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You are right, so it turns out that the error in the book is that it should have asked for the minimum static friction, not the kinetic friction, and by coincidence both "error corrected" versions of the question have the same answer.
    By the way, there is an error in one line of the textbook solution as you have posted it. It should read ##mg\sin(37)-\mu mg\cos(37)##.
  8. Sep 9, 2016 #7
    Yeah, that was the same thing written in the book its a mistake on my side, it ought to be ##mg\sin{37^{\circ}}-\mu mg\cos{37^{\circ}}=m\times(5/3)##. Also, according to the FBD the book has drawn doesn't the force equation in the direction of the acceleration look faulty instead of ##mg\sin{37^{\circ}}-\mu mg\cos{37^{\circ}}=m\times(5/3)## shouldn't it be ##mg\sin{37^{\circ}}+\mu mg\cos{37^{\circ}}=m\times(5/3)##
    Last edited: Sep 9, 2016
  9. Sep 9, 2016 #8
    I don't think the book's answer is correct regarding the coefficient of static friction b/w the platform and the man because it has not included the normal's component along the direction of the acceleration of the man.
    The expression should be
    $$mg\sin{37^{\circ}}+\mu_s N_1\cos{37^{\circ}}-N_1\sin{37^{\circ}}=m\times(5/3) \\
    \implies (80-72)g\sin{37^{\circ}}+\mu_s 72g\cos{37^{\circ}}=80\times(5/3) \\
    \implies \mu_s=\dfrac{4}{27}$$

    or the simpler method that I did in the first post
    $$\mu_s N_1=ma_P\cos{37^{\circ}}\implies \mu_s=\dfrac{4}{27}$$
  10. Sep 9, 2016 #9


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, sorry, I forgot that part... looking at too many different threads. So the book is completely wrong.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Problem with friction
  1. A problem on friction (Replies: 8)

  2. Friction Problem (Replies: 2)

  3. Friction problem (Replies: 5)

  4. Friction Problem (Replies: 13)