# 10.6 Induced motion on rough incline plane  (Page 3/3)

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The components of forces parallel to contact surface (excluding friction) is :

$\begin{array}{l}\sum {F}_{x}=F\mathrm{cos}{30}^{0}-mg\mathrm{sin}{45}^{0}\\ ⇒{F}_{\mathrm{net}}=10\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}\frac{\surd 3}{2}-5\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}10\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}\frac{1}{\surd 2}=-26.7\phantom{\rule{2pt}{0ex}}N\\ ⇒{F}_{\mathrm{net}}<0\end{array}$

It means that the block has tendency to move in the downward direction. The friction, therefore, is in up direction. Now, this completes the force system on the block as shown here.

An additional external force with component parallel to contact up the incline, does not guarantee upward motion. The force of gravity can still be greater than the component of applied additional external force parallel to the contact surface.

Now, our next task is to know the state of friction. To know the state of friction, we need to compare the net force component parallel to contact with maximum static friction. Now, maximum friction force is :

$\begin{array}{l}{F}_{s}={\mu }_{s}N=0.55N\end{array}$

In order to evaluate maximum static friction, we need to know normal force on the block. It is clear from the free body diagram that we can find normal force, "N" by analyzing force in y-direction. As there is no motion in vertical direction, the components in this direction form a balanced force system.

$\begin{array}{l}\sum {F}_{y}=N+F\mathrm{sin}{30}^{0}-mg\mathrm{cos}{45}^{0}=0\\ ⇒N=mg\mathrm{cos}{45}^{0}-F\mathrm{sin}{30}^{0}\\ ⇒N=5x10x\frac{1}{\surd 2}-10x\frac{1}{2}\\ ⇒N=30.36\phantom{\rule{2pt}{0ex}}\mathrm{Newton}\end{array}$

The maximum friction force is :

$\begin{array}{l}⇒{F}_{s}={\mu }_{s}N=0.55x30.36=16.7\phantom{\rule{2pt}{0ex}}N\end{array}$

Thus magnitude of net component parallel to contact surface (26.7) is greater than maximum static friction (16.7 N). Thus, body will slide down and friction will be equal to kinetic friction. Hence,

$\begin{array}{l}⇒{F}_{F}={F}_{k}={\mu }_{k}N=0.53x30.36=16.1\phantom{\rule{2pt}{0ex}}N\end{array}$

## Case 2 : friction equal to maximum static friction

In this case, net force parallel to contact surface is equal to maximum static friction.

$\begin{array}{l}{F}_{F}={F}_{s}={\mu }_{s}N\end{array}$

Problem : A block of 10 kg rests on a rough incline of angle 45°. The block is tied to a horizontal string as shown in the figure. Determine the coefficient of friction between the surfaces, if tension in the string is 50 N. Consider g = 10 $\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ .

Solution : Here, block is stationary. We need to know the motional tendency, had the string were not there.

It is given that the string is taught with a tension of 50 N. Clearly, the block would have moved down had it not been held by the string is taut. The direction of friction, therefore, is upwards. Further, the system is static with taut string. It means that friction has reached maximum static friction. Otherwise, how would have block moved down if not held by the string. Hence, the forces on the block form a balanced force system, including maximum static friction acting upward.

$\begin{array}{l}\sum {F}_{x}=mg\mathrm{sin}{45}^{0}-{\mu }_{s}N-T\mathrm{cos}{45}^{0}=0\\ ⇒{\mu }_{s}N=10x10x\frac{1}{\surd 2}-50x\frac{1}{\surd 2}=50\frac{1}{\surd 2}\end{array}$

and

$\begin{array}{l}\sum {F}_{y}=N-mg\mathrm{cos}{45}^{0}-T\mathrm{sin}{45}^{0}=0\\ ⇒N=10x10x\frac{1}{\surd 2}+50x\frac{1}{\surd 2}=150\frac{1}{\surd 2}\end{array}$

$\begin{array}{l}⇒{\mu }_{s}=\frac{1}{3}\end{array}$

## Case 3 : all friction states

Problem : A block attached with a spring (of spring constant “k”) is placed on a rough horizontal plate. The spring is in un-stretched condition, when plate is horizontal. The plate is, then, raised with one end gradually as shown in the figure. Analyze friction and extension of the spring as angle (θ) increases to 90°.

Solution : In the initial stages, spring does not apply any force as block remains stationary because of friction. The friction acts in up direction as there is no applied external force due to spring. The magnitude of friction is equal to the component of weight along the incline. The friction and extension in the spring (x) for different ranges of angle “θ” are given here. Note that ${\mathrm{tan}}^{-1}\left({\mu }_{s}\right)$ is equal to angle of repose. This is the angle of inclination of the plate with horizon, when block begins to move down.

(i) For $\theta <{\mathrm{tan}}^{-1}\left({\mu }_{s}\right)$

$\begin{array}{l}⇒{f}_{s}=mg\mathrm{sin}\theta \phantom{\rule{2pt}{0ex}};\phantom{\rule{2pt}{0ex}}x=0\end{array}$

(ii) For $\theta ={\mathrm{tan}}^{-1}\left({\mu }_{s}\right)$

Till the angle does not reach the value equal to angle of repose, there is no motion of the block.

$\begin{array}{l}⇒{F}_{s}={\mu }_{s}mg\mathrm{cos}\theta \phantom{\rule{2pt}{0ex}};\phantom{\rule{2pt}{0ex}}x=0\end{array}$

(iii) For $\theta >{\mathrm{tan}}^{-1}\left({\mu }_{s}\right)$

The block starts sliding down. The direction of friction is up along the incline. As the spring stretches, it applies spring force to counteract the net downward force to again bring the block to rest. Before, the block is brought to a stop, the acceleration of the block (“a”) is :

$\begin{array}{l}\sum {F}_{x}=mg\mathrm{cos}\theta -{\mu }_{s}mg\mathrm{cos}\theta -kx=ma\\ ⇒a=\frac{mg\mathrm{cos}\theta -{\mu }_{s}mg\mathrm{cos}\theta -kx}{m}\end{array}$

When the block is brought to a stop, the extension in the string ${x}_{0}$ is obtained as under :

$\begin{array}{l}\sum {F}_{x}=mg\mathrm{cos}\theta -{\mu }_{s}mg\mathrm{cos}\theta -k{x}_{0}=0\\ ⇒k{x}_{0}=mg\mathrm{sin}\theta -{\mu }_{k}mg\mathrm{cos}\theta \\ ⇒{x}_{0}=\frac{mg\mathrm{sin}{\theta }_{0}-{\mu }_{s}mg\mathrm{cos}{\theta }_{0}}{k}\end{array}$

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