# Question #a8cb9

Dec 4, 2016

${\mu}_{s} \left(7.5 \times 9.8 - 18 \sin {38}^{\circ}\right) N$

#### Explanation:

$F \to \text{Applied force } = 18 N$

$\alpha \to \text{angle of applied force with the horizontal } = {38}^{\circ}$

${F}_{v} \to \text{vertical component of applied force} = F \sin \alpha$

${F}_{h} \to \text{horizontal component of applied force} = F \cos \alpha$

$m \to \text{mass of the object} = 7.5 k g$

$N \to \text{Normal reaction} = m g - F \sin \alpha$

${F}_{\text{fric"->"Limiting value of friction}} = {\mu}_{s} \times N$

If horizontal component of the applied force represents the limiting value of the frictional force then

${F}_{\text{fric}} = {F}_{h}$

$\implies {\mu}_{s} \times N = F \cos \alpha$

$\implies {\mu}_{s} = \frac{F \cos \alpha}{N} = \frac{F \cos \alpha}{m g - F \sin \alpha}$

$= \frac{18 \times \cos 38}{7.5 \times 9.8 - 18 \times \sin 38} \approx 0.23$