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

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 $m/s^2$ .

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.

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

and

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

$$\begin{array}{l}\Rightarrow {\mu }_s=\frac{1}{3}\end{array}$$

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 ${\tan }^{-1}\left({\mu }_s\right)$ 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 $\theta <{\tan }^{-1}\left({\mu }_s\right)$

$$\begin{array}{l}\Rightarrow f_s=mg\sin \theta ;x=0\end{array}$$

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

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

$$\begin{array}{l}\Rightarrow F_s={\mu }_{s}mg\cos \theta ;x=0\end{array}$$

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

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 :

$$\begin{array}{l}\sum F_x=mg\cos \theta -{\mu }_{s}mg\cos \theta -kx=ma\\ \Rightarrow a=\frac{mg\cos \theta -{\mu }_{s}mg\cos \theta -kx}{m}\end{array}$$

When the block is brought to a stop, the extension in the string $x_0$ is obtained as under :

$$\begin{array}{l}\sum F_x=mg\cos \theta -{\mu }_{s}mg\cos \theta -k{x}_0=0\\ \Rightarrow k{x}_0=mg\sin \theta -{\mu }_{k}mg\cos \theta \\ \Rightarrow x_0=\frac{mg\sin {\theta }_0-{\mu }_{s}mg\cos {\theta }_0}{k}\end{array}$$