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 _____ respectively.


Answer:$F$,$N$ and $f_L$

I need an explanaton

  • $\begingroup$ Just examine the end points. At $\theta = 0$, flat plane, what would you expect the net force to be? At $\theta = 90$, vertical plane, what would you expect the normal to be? Etc... $\endgroup$ – npojo Mar 16 '18 at 10:38

To help you decide which graph is which I have added the graph of component of the weight of the block down the slope.

enter image description here

Consider how the component of the weight of the block is related to the angle of inclination $\theta$ and also how the normal reaction force relates to the angle $\theta$.

Remember that the limiting friction force is proportional to the normal reaction with the constant of proportionality being the coefficient of static friction $\mu_{\rm static}$ which is usually less than one.
The actual static frictional force acting on the block will be less than the limiting frictional force until the block is on the point of moving.

Once the block moves relative to the inclined plane the kinetic frictional force replaces the static frictional force and will be less than the component of the weight down the inclined slope.


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