Question

In the situation shown in the figure, block of mass $m$ is pulled so that the spring is elongated without the block slipping. Which of the following graphs correctly represents the relation between static frictional force and elongation for angle of inclination ${37}^{\circ }$ and ${53}^{\circ }$. The spring constant is $k$ and the coefficient of friction is ${\mu }_s=0.8$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

FBD of block for ${37}^{\circ }$ inclination is as shown


From the FBD, we have
$R=mg\cos {37}^{\circ }$
$mg\sin {37}^{\circ }+kx=f-----\left(1\right)$

Since $f=\mu R$ in limiting case, we get
$kx=\mu \left(mg\cos {37}^{\circ }\right)-mg\sin {37}^{\circ }$
$x=\frac{mg}{k}(\mu \cos {37}^{\circ }-\sin {37}^{\circ })$
$x>0$
$\Rightarrow \mu \cos {37}^{\circ }-\sin {37}^{\circ }>0$
$\Rightarrow \mu \cos {37}^{\circ }>\sin {37}^{\circ }$
$\Rightarrow \tan {37}^{\circ }<\mu$
$\Rightarrow \tan {37}^{\circ }=\frac{3}{4}<\mu =0.8$
Thus, the block will not slip.
Since eq. (1) looks like $y=mx+c$
$f$ vs $x$ is a straight line

Now, FBD of the block for ${53}^{\circ }$ inclination is as shown.


From the FBD, we have
$R=mg\cos {53}^{\circ }$
$mg\sin {53}^{\circ }+kx=f----\left(2\right)$

Since $f=\mu R$ in limiting case, we get
$kx=\mu \left(mg\cos {53}^{\circ }\right)-mg\sin {53}^{\circ }$
$x=\frac{mg}{k}(\mu \cos {53}^{\circ }-\sin {53}^{\circ })$
$x>0$
$\Rightarrow \mu \cos {53}^{\circ }-\sin {53}^{\circ }>0$
$\Rightarrow \mu \cos {53}^{\circ }>\sin {53}^{\circ }$
$\Rightarrow \tan {53}^{\circ }<\mu$
But $\tan {53}^{\circ }=\frac{4}{3}>\mu =0.8$
$\therefore$ Block will slip at this angle
Hence, static friction $f=0$