Question

In Figure 15-37, two blocks$(m=1.8\mathrm{kg}\mathrm{and}M=10\mathrm{kg})$(and) and a spring (k=200 N/m) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.40.What amplitude of simple harmonic motion of the spring–blocks system puts the smaller block on the verge of slipping over the larger block?

Short answer

The amplitude of SHM of the spring-blocks system that puts the smaller block on the verge of slipping over the larger block is 0.23 m.

Step-by-step solution

The given data

1. Masses of the blocks,$\mathrm{m}=1.8\mathrm{kg},\mathrm{M}=10\mathrm{kg}$
2. Spring constant, $\mathrm{k}=200\mathrm{N}/\mathrm{m}$
3. The coefficient of static friction between two blocks,${\mathrm{\mu }}_{\mathrm{s}}=0.40$

Understanding the concept of Frictional force and motion

We write for the frictional force between two blocks from the condition for the smaller block which is on the verge of slipping. Then inserting the formulae for frictional force and angular frequency, we can find the amplitude of SHM of the spring-blocks system that puts the smaller block on the verge of slipping over the larger block.

Formula:

Newton’s second law gives the force,${\mathrm{F}}_{\mathrm{net}}=\mathrm{ma}$………(i)

Static frictional force,${\mathrm{F}}_{\mathrm{net}}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{mg}$……….(ii)

Maximum acceleration of SHM,${\mathrm{a}}_{\mathrm{m}}={\mathrm{\omega }}^2{\mathrm{X}}_{\mathrm{m}}$……….(iii)

Calculation of amplitude of SHM

Smaller block will be on the verge slipping over the larger block if the net force acting on it will be zero. For the condition of friction, the net force should be equal to that of the static frictional force. Hence, using equations (i) & (ii), we get the angular frequency as follows:

${\mathrm{\mu }}_{\mathrm{S}}\mathrm{mg}={\mathrm{ma}}_{\mathrm{m}}$………..(iv)

Substitute the value of ${\mathrm{a}}_{\mathrm{m}}$

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{s}}\mathrm{mg}={\mathrm{m\omega }}^2{\mathrm{x}}_{\mathrm{m}} \\ \mathrm{\omega }=\sqrt{\frac{\mathrm{k}}{\mathrm{m}+\mathrm{M}}}(\mathrm{Angular}\mathrm{frequency}=\mathrm{\omega }) \end{gathered}$$

Hence, using the above derived& given values in equation (iii), we get the amplitude as

follows:

$$\begin{gathered} {\mathrm{x}}_{\mathrm{m}}=\frac{{\mathrm{\mu }}_{\mathrm{s}}\mathrm{g}\left(\mathrm{m}+\mathrm{M}\right)}{\mathrm{k}} \\ =\frac{(0.40)\left(9.8\right)\left(1.8+10\right)}{200} \\ =0.23\mathrm{m} \\ =23\mathrm{cm} \end{gathered}$$ $\left(\mathrm{from}\mathrm{equation}(\mathrm{iv}\right),\mathrm{we}\mathrm{get}{\mathrm{a}}_{\mathrm{m}})$

Therefore, the amplitude of SHM of the spring-blocks system puts the smaller block on the verge of slipping over the larger block is 0.23m.