Question

In Fig. 12-64, block A $(mass10kg)$is in equilibrium, but itwould slip if block B $(mass5.0kg)$were any heavier. For angle $\text{θ}=30\text{°}$what is the coefficient of static friction between block Aand the surfacebelow it?

Short answer

The coefficient of static friction between block A and the surface is$0.288$

Step-by-step solution

Listing the given quantities

${\text{m}}_{\text{A}}=10kg$

${\text{m}}_{\text{B}}=5kg$

$\theta =30°$

Understanding the concept of coefficient of static friction

By drawing the free body diagram of the situation, we can find the tension in each wire employing which they are connected. Using the conditions for equilibrium, we can find the coefficient of static friction.

Formula:

${\sum }_{}^{}{\text{F}}_{\text{y}}=0$

${\sum }_{}^{}{\text{F}}_{\text{x}}=0$

$F_{staticfriction}={\text{μ}}_{\text{s}}\times NormalForce$

Free body diagram block A

From this, we can say that,

${\text{T}}_{\text{1}}={\text{F}}_{\text{s}}$

${\text{T}}_{\text{1}}={\text{μ}}_{\text{s}}\times {\text{F}}_{\text{N}}$

${\text{T}}_{\text{1}}={\text{μ}}_{\text{s}}\times ({\text{m}}_{\text{A}}\times \text{g})$

${\text{T}}_{\text{1}}={\text{μ}}_{\text{s}}\times (10\text{\hspace{0.17em}kg}\times 9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})$

${\text{T}}_{\text{1}}=98\text{\hspace{0.17em}N}\times {\text{μ}}_{\text{s}}$

Free body diagram block B

From this, we can say that,

${\text{T}}_{\text{2}}={\text{m}}_{\text{B}}\times \text{g}$

$\begin{array}{l}{T}_2=5kg\times 9.8m/s^2\\ T_2=49N\end{array}$

Free body diagram for equilibrium position

Calculations of static friction

From this, we can say that,

$\begin{array}{l}{\sum }_{}^{}{\text{F}}_{\text{y}}=0\\ \left({\text{T}}_{\text{3}}\text{×cosθ}\right){\text{-T}}_{\text{2}}=0\\ \left({\text{T}}_{\text{3}}\text{×cosθ}\right)={\text{T}}_{\text{2}}\\ \left({\text{T}}_{\text{3}}\text{×cosθ}\right)=49N\end{array}$

$\begin{array}{l}{\sum }_{}^{}{\text{F}}_{\text{x}}=0\\ ({\text{T}}_{\text{3}}\times \text{sinθ}){\text{-T}}_{\text{1}}=0\\ ({\text{T}}_{\text{3}}\times \text{sinθ})={\text{T}}_{\text{1}}\\ ({\text{T}}_{\text{3}}\times \text{sinθ})=98\text{\hspace{0.17em}N}\times {\mu }_{\text{s}}\end{array}$

$\begin{array}{l}\frac{({\text{T}}_{\text{3}}\times \text{sinθ})}{({\text{T}}_{\text{3}}\times \text{cosθ})}=\frac{98N\times {\text{μ}}_{\text{s}}}{49N}\\ {\text{μ}}_{\text{s}}=\tan \theta \times 0.5\\ {\text{μ}}_{\text{s}}=0.5\times \tan (30°)\\ {\text{μ}}_{\text{s}}=0.5\times 0.577\\ {\text{μ}}_{\text{s}}=0.288\end{array}$

The coefficient of static friction between block A and the surface is$0.288$