Due to friction ,the system as shown in the diagram remains motionless. Calculate the static coefficient of friction ?

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Due to friction ,the system as shown in the diagram remains motionless. Calculate the static coefficient of friction ?

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Answer 1 · 2017-08-16T15:39:33

$μ_{s} \leq 0.346$ $mu_s<=0.346$

Explanation:

Here's what I tried.

I defined up the ramp as the positive direction.

Forces on $m_{1}$ $m_1$ :

$\sum F_{x} = T_{1} - {(F_{G})}_{x} - f_{s} = 0$ $sumF_x=T_1-(F_G)_x-f_s=0$

$\sum F_{y} = n - {(F_{G})}_{y} = 0$ $sumF_y=n-(F_G)_y=0$

$f_{s max} = μ_{s} n$ $f_(s"max")=mu_sn$
${(F_{G})}_{x} = m_{1} g sin (θ)$ $(F_G)_x=m_1gsin(theta)$
${(F_{G})}_{y} = m_{2} g cos (θ)$ $(F_G)_y=m_2gcos(theta)$
$n = m g cos (θ)$ $n=mgcos(theta)$

I will refer to $f_{s max}$ $f_(s"max")$ simply as $f_{s}$ $f_s$ from this point on, though I am still solving in terms of the maximum static friction.

$\Rightarrow f_{s} = T_{1} - m_{1} g sin (θ)$ $=>f_s=T_1-m_1gsin(theta)$

$\Rightarrow μ_{s} m_{1} g cos (θ) = T_{1} - m_{1} g sin (θ)$ $=>mu_sm_1gcos(theta)=T_1-m_1gsin(theta)$

$\Rightarrow μ_{s} = \frac{T_{1} - m_{1} g sin (θ)}{m_{1} g cos (θ)}$ $=>color(darkblue)(mu_s=(T_1-m_1gsin(theta))/(m_1gcos(theta)))$

Forces on $m_{2}$ $m_2$ :

$\sum F = \sum F_{y} = T_{2} - F_{G} = 0$ $sumF=sumF_y=T_2-F_G=0$

$F_{G} = m_{2} g$ $F_G=m_2g$

Because we can assume a massless rope and frictionless pulley, ${\to T}_{1}$ $vecT_1$ and ${\to T}_{2}$ $vecT_2$ act as an "action/reaction" pair.

$T_{1} = T_{2} = m_{2} g$ $T_1=T_2=m_2g$

$\Rightarrow μ_{s} = \frac{m_{2} g - m_{1} g sin (θ)}{m_{1} g cos (θ)}$ $=>mu_s=(m_2g-m_1gsin(theta))/(m_1gcos(theta))$

$=>mu_s=(cancel(g)(m_2-m_1sin(theta)))/(cancel(g)(m_1cos(theta))$

$=>color(darkblue)(mu_s=(m_2-m_1sin(theta))/(m_1cos(theta)))$

Using known values:

$mu_s=(80-100(0.500))/(100(0.866))$

$mu_(s"max")=0.346$

$=>mu_s<=0.346$

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Due to friction ,the system as shown in the diagram remains motionless. Calculate the static coefficient of friction ?

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