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

Aug 16, 2017

${\mu}_{s} \le 0.346$

#### Explanation:

Here's what I tried.

• I defined up the ramp as the positive direction.

Forces on ${m}_{1}$:

$\sum {F}_{x} = {T}_{1} - {\left({F}_{G}\right)}_{x} - {f}_{s} = 0$

$\sum {F}_{y} = n - {\left({F}_{G}\right)}_{y} = 0$

• ${f}_{s \text{max}} = {\mu}_{s} n$

• ${\left({F}_{G}\right)}_{x} = {m}_{1} g \sin \left(\theta\right)$

• ${\left({F}_{G}\right)}_{y} = {m}_{2} g \cos \left(\theta\right)$

• $n = m g \cos \left(\theta\right)$

I will refer to ${f}_{s \text{max}}$ simply as ${f}_{s}$ from this point on, though I am still solving in terms of the maximum static friction.

$\implies {f}_{s} = {T}_{1} - {m}_{1} g \sin \left(\theta\right)$

$\implies {\mu}_{s} {m}_{1} g \cos \left(\theta\right) = {T}_{1} - {m}_{1} g \sin \left(\theta\right)$

$\implies \textcolor{\mathrm{da} r k b l u e}{{\mu}_{s} = \frac{{T}_{1} - {m}_{1} g \sin \left(\theta\right)}{{m}_{1} g \cos \left(\theta\right)}}$

Forces on ${m}_{2}$:

$\sum F = \sum {F}_{y} = {T}_{2} - {F}_{G} = 0$

• ${F}_{G} = {m}_{2} g$

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

• ${T}_{1} = {T}_{2} = {m}_{2} g$

$\implies {\mu}_{s} = \frac{{m}_{2} g - {m}_{1} g \sin \left(\theta\right)}{{m}_{1} g \cos \left(\theta\right)}$

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

$\implies \textcolor{\mathrm{da} r k b l u e}{{\mu}_{s} = \frac{{m}_{2} - {m}_{1} \sin \left(\theta\right)}{{m}_{1} \cos \left(\theta\right)}}$

Using known values:

${\mu}_{s} = \frac{80 - 100 \left(0.500\right)}{100 \left(0.866\right)}$

${\mu}_{s \text{max}} = 0.346$

$\implies {\mu}_{s} \le 0.346$