Question

A block of mass $m$ is placed on an inclined plane with angle of inclination $\theta$. Let $N,f_L$ and $F$ respectively represent the normal reaction, limiting force of friction and the net force down the inclined plane. Let $\mu$ be the coefficient of friction. The dependence of $N,f_L$ and $F$ on $\theta$ is indicated by plotting graphs as shown below. Then curves (1), (2) and (3) respectively represent

• A. $N,Fand{f}_L$
• B. $f_L,FandN$
• C. $F,Nand{f}_L$
• D. $f_L,NandF$

Answer 1

Answer: (C)

$F,Nand{f}_L$

The normal reaction is $N=mgcos\theta$. $cos\theta$ will vary from 1 to 0 as the $\theta$ vary from 0 to $\pi$. So the N will be represented by curve (2).

Now, $f_L=\mu N$ and $\mu <1$, so the $f_L$ will follow all the same trend as N with a reduced amplitude. So the curve no (3).

The net downward force is given by:

$F=(mgsin\theta -f_L)$. When the block will cross the limiting friction then it will start moving and the frictional force will be constant. So the second term in the bracket will be constant. So $f_L$ will follow the nature of a $sin\theta$ curve with a origin shift to negative y-axis. So the curve no (1)