Question

A block is limiting equilibrium on a rough horizontal surface. If the net contact force is $\sqrt{3}$ times the normal force, the coefficient of the static friction is

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $0.5$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is B

$\sqrt{2}$

Step 1: Given data

Net contact force is $\sqrt{3}$ times the normal force.

Step 2: Formula and definition

Contact force is any force that requires contact and is often decomposed into its orthogonal components.

Here the orthogonal components are normal force and frictional force.

We know, that the magnitude of the square of a vector is denoted by the sum of squares of the magnitudes of its components.

So, ${(netcontactforce)}^2=(normalforce{)}^2+{(frictionalforce)}^2$

$f_{\mathit{l}}\mathit{=}{\mu }_{\mathit{s}}N\mathit{}\mathit{}\mathit{}\mathit{(}{f}_{\mathit{l}}\mathit{=}limiting\mathit{}friction\mathit{}force\mathit{,}\mathit{}{\mu }_{\mathit{s}}\mathit{=}static\mathit{}friction\mathit{}cofficient\mathit{}\mathit{,}\mathit{}N\mathit{=}normal\mathit{}force\mathit{)}$

Step 3: Calculating the coefficient of static friction

Let us assume the normal force is $\mathrm{N}$ (acting vertically upward) and the coefficient of static friction = $\mu$

So the limiting friction is $f_l=\mu N$ (acting horizontal)

$f_c$ is the net contact force and $f_n$ is the net normal force.

The direction of $f_c$ is the resultant of $f_n$ and $f_i$.

Therefore, the net contact force is

${(netcontactforce)}^2=(normalforce{)}^2+{(frictionalforce)}^2$

$$\begin{gathered} f_c^2=f_l^2+f_n^2 \\ \Rightarrow f_c=\sqrt{f_l^2+f_n^2} \\ \Rightarrow f_c=\sqrt{{\left(\mu N\right)}^2+N^2} \\ \Rightarrow f_c=\sqrt{N^2{(\mu }^2+1)} \\ \Rightarrow f_c=N\sqrt{{1+\mu }^2} \end{gathered}$$

The static friction = horizontal force acting on the block (given)

$$\begin{gathered} N\sqrt{(1+{\mu }^2)}=\sqrt{3}N \\ \Rightarrow \sqrt{(1+{\mu }^2)}=\sqrt{3} \\ \Rightarrow (1+{\mu }^2)=3 \\ \Rightarrow {\mu }^2=2 \\ \Rightarrow \mu =\sqrt{2} \end{gathered}$$

Thus the coefficient of static friction is $\sqrt{2}$.

Hence, option B is the correct answer.