- #1
decentfellow
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Homework Statement
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.
Calculate:
(a) The vertical acceleration of the man.
(b) The coefficient of kinetic friction ##\mu## between the platform and man.
Homework Equations
$$\begin{align}
&f_k\le \mu_{k}mg\qquad\qquad \text{where, $\mu_{k}\rightarrow$ coefficent of static friction} \\
&f_s\le \mu_{s}mg\qquad\qquad \text{where, $\mu_{s}\rightarrow$ coefficent of static friction} \\
&F=ma
\end{align}$$
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.
$$m_Mg-N_1=m_Ma_P\sin{37^{\circ}}\qquad\qquad\text{where, $m_M$ is the mass of man} \\
\implies (80-72)g=80a_P\left(\dfrac{3}{5}\right)\implies a_P=\dfrac{5}{3}\text{m/s$^2$}$$
So, from the above FBD of the man standing on the platform we see that
$$f_s=m_Ma_P\cos\theta$$
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.
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