In Fig. 6-37, a slab of mass ${}^{{\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{40}\mathbf{}\mathbf{k}\mathbf{g}}$rests on a frictionless floor, and a block of mas ${}^{{\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{k}\mathbf{g}}$rests on top of the slab. Between block and slab, the coefficient of static friction is ${}^{\mathbf{0}\mathbf{.}\mathbf{60}}$, and the coefficient of kinetic friction is ${}^{\mathbf{0}\mathbf{.}\mathbf{40}}$. A horizontal force of magnitude ${}^{\mathbf{100}\mathbf{N}}$begins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of (a) the block and (b) the slab?

$$\begin{gathered} m_1=40\mathrm{kg} \\ {m}_2=10\mathrm{kg} \end{gathered}$$

Coefficient of static friction between block and slab = 0.60,

Coefficient of kinetic friction is 0.40,

Horizontal force =100 N.

Use the concept of friction and Newton's first law of motion. According to Newton’s first law, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

Formula:

$F_{net}=\sum _{}ma$

where, F is the net force, m is mass and a is an acceleration.

The free-body diagrams for the slab and block are shown below:

F is the 100 N force applied to the block, ${}^{F_{Ns}}$is the normal force of the floor on the slab,

${}^{F_{Nb}}$is the magnitude of the normal force between the slab and the block, f is the force of friction between the slab and the block, $m_s$is the mass of the slab, and $m_b$is the mass of the block. For both objects, take the +x direction to the right and the +y direction to be up.

Applying Newton’s second law for the x and y axes for (first) the slab and (second) the block results in four equations:

$$\begin{gathered} -f=m_x{a}_x \\ {F}_{Ns}-F_{Nb}-m_{s}g=0 \\ f-F=m_b{a}_b \\ {F}_{Nb}-m_{b}g=0 \end{gathered}$$

From which, note that the maximum possible static friction magnitude would be,

$$\begin{gathered} {\mu }_s{F}_{Nb}={\mu }_s{m}_{b}g \\ =\left(0,6\right)\left(10\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =59\mathrm{N} \end{gathered}$$

Check to see if the block slides on the slab. Assuming, it does not, then localid="1654235514304" ${}^{a_s=a_b}$(which we denote simply as a) and solve for localid="1654235535303" ${}^f$

$$\begin{gathered} f=\frac{m_{s}F}{m_s+m_b} \\ =\frac{\left(40\mathrm{kg}\right)\left(100\mathrm{N}\right)}{40\mathrm{kg}+10\mathrm{kg}} \\ =80\mathrm{N} \end{gathered}$$

Which is greater than$f_s$,max so that the block is sliding across the slab (their accelerations are different).

Using,$f={\mu }_k{F}_{Nb},$the above equations yield,

$$\begin{gathered} a_b=\frac{{\mu }_k{m}_{b}g-F}{m_b} \\ =\frac{\left(0.40\right)\left(10\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-100\mathrm{N}}{10\mathrm{kg}} \\ =-6.1\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

T negative sign means that the acceleration is leftward. That is, $a_b=\left(-6.1\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}$

Hence, the resulting accelerations of the block is$\left(-6.1\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}$

Also,

$$\begin{gathered} a_s=-\frac{{\mu }_k{m}_{b}g}{m_s} \\ =-\frac{\left(0.40\right)\left(10\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{40\mathrm{kg}} \\ =-0.98\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

As mentioned above, this means it accelerates to the left. That is,$a_s=\left(-0.98\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}$

Hence, the resulting accelerations of the slab is$\left(-0.98\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}$