Question

The two blocks in FIGURE P15.52 oscillate on a frictionless surface with a period of $1.5s$. The upper block just begins to slip when the amplitude is increased to $40cm$. What is the coefficient of static friction between the two blocks?

FIGURE P15.52

Short answer

The coefficient of static friction between the two blocks is${\mu }_s=0.72$

Step-by-step solution

 Step 1 : Introduction 

The given is the period of oscillation of two blocks $T=1.5s$and the upper block slips when the amplitude is $A=40cm$. The objective is to find coefficient of static friction between the two blocks. Newton's Second Law and Static Friction Force is used to solve

:

Newton's Second Law states that the net force F acting on a mass m body is proportional to the acceleration a.

$\sum _{}f=ma$ .................(1)

:

When two things are in contact with one other and we try to move one across the other, they exert forces on each other. Static friction forces oppose the tendency of the objects to move with regard to one other, and are the components of these forces parallel to the surface. The maximum static friction force is proportional to the size of the normal force n applied on one surface on the other, and it is determined by the roughness of the surfaces of the two objects (the coefficient of static friction ${\mu }_s$between the surfaces).

${\mu }_{s,max}={\mu }_{s}n$ ..............(2)

Step 4:

In terms of angular speed $\omega$and amplitude A, the maximum transverse acceleration of a particle in simple harmonic motion is calculated as follows:

$a_{max}={\omega }^{2}A$ ...............(3)

:

In simple harmonic motion, the angular frequency of an oscillator is given by

$\omega =\sqrt{\frac{k}{m}}$ ................(4)

Note that depends solely on the mass m and the force constant k, not on the amplitude.

:

The amount of the static friction force between the two blocks fluctuates until it reaches its maximum values just as the upper block is about to slip.

The forces operating on the higher block are shown in Figure 1:$\overrightarrow{n}$is the normal force exerted by the lower block,$m\overrightarrow{g}$is the size of the earth's gravitational pull, and${\overrightarrow{f}}_{s,max}$is the magnitude of the greatest friction force between the two blocks.

:

Because the higher block does not move in the vertical direction, the vertical net force on it must be zero.

$F_g=n-mg=0$

$n=mg$ ...............(5)

Step 8:

We get Equation (1) by applying Newton's second law to the upper block in the horizontal direction.

$f_x=f_{s,max}=m{a}_{x,max}$

Substitute for localid="1650092848100" $f_{s,max}$from equation $\left(2\right)$and for localid="1650092854289" $a_{x,max}$from the equation $\left(3\right)$where a maximum value of acceleration, max corresponds to the maximum value of friction. :

${\mu }_{s}n=m{\omega }^{2}A$

Substitute n and localid="1650092973988" $\omega$from equations (5) and (4) respectively localid="1650092952928" $\left(4\right)$

${\mu }_{s}mg=m{\left(\frac{2\mathrm{\pi }}{T}\right)}^{2}A$

${\mu }_{s}g={\left(\frac{2\mathrm{\pi }}{T}\right)}^{2}A$

solve for${\mu }_s$

${\mu }_s=\frac{{\left({\displaystyle \frac{2\mathrm{\pi }}{T}}\right)}^{2}A}{g}$localid="1650092940545" ${\mu }_s$

localid="1650092839334" ${\mu }_s=\frac{{\left({\displaystyle \frac{2\mathrm{\pi }}{T}}\right)}^{2}A}{g}$

Substitute the numerical values

${\mu }_s=\frac{{\left({\displaystyle \frac{2\mathrm{\pi }}{1.5s}}\right)}^2\times (0.4m)}{9..80{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}$

${\mu }_s=0.72$