1. Write a realistic problem for which these are the correct equations.
2. Draw the free-body diagram and the pictorial representation for your problem.
3. Finish the solution of the problem.

$T-0.20n-\left(20\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)\sin {20}^{\circ }$

$\begin{array}{rl}& =\left(20\mathrm{kg}\right)(2.0\mathrm{m}/{\mathrm{s}}^2)\\ n& -\left(20\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)\cos {20}^{\circ }=0\end{array}$

Two given equations are:

$\begin{array}{rl}& -0.80n=\left(1500\mathrm{kg}\right)a_x\\ & n-\left(1500\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)=0\end{array}$

In the above equations, $n$and $a_x$can be represented by the normal force applied to the upper block and acceleration of the lower block.

The first equation represents the equilibrium between the frictional force between the surface and the applied force on the lower block. If the applied force is greater than the frictional force between the surface, then the upper block will slide. Knowing $n$from the second equation, $a_x$ is calculated from this equation.

The second equation represents the equilibrium between the weight of the upper block and the normal reaction on the upper block. From this equation, we can calculate the normal reaction.

Two given equations are:

$\begin{array}{rl}& -0.80n=\left(1500\mathrm{kg}\right)a_x\\ & n-\left(1500\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)=0\end{array}$

In the free body diagram, $F$represents the force applied on the block by the lower block, and $f_f$represents the frictional force applied to the upper block. The other two forces are$w=mg$is the weight of the block and $n$is the normal reaction, which is equal to the weight of the block. Since the block is in equilibrium, we must have

$F=f_f$

And

$n=w.$

$\begin{array}{rl}& -0.80n=\left(1500\mathrm{kg}\right)a_x\\ & n-\left(1500\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)=0.\end{array}$

Solve second equation to calculate $n$.

$\begin{array}{c}n-\left(1500\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)=0\\ \Rightarrow n=\left(1500\mathrm{kg}\right)(9.80\mathrm{m}/{\mathrm{s}}^2)\\ \Rightarrow n=14700\mathrm{kg}\end{array}$

Now, substitute of the value of $n$in the first equation to calculate the value of $a_x$.

$\begin{array}{c}-0.80n=\left(1500\mathrm{kg}\right)a_x\\ \Rightarrow a_x=-\frac{0.80\mathrm{n}}{1500\mathrm{kg}}=-\frac{0.80\left(14700\mathrm{N}\right)}{1500\mathrm{kg}}=-7.84\mathrm{m}/{\mathrm{s}}^2\end{array}$

Part (a): A block of mass $1500\mathrm{kg}$is kept on another block that is moving with acceleration $a_x$. The coefficient of static friction between the surface of the block is $\mu =0.80$. Calculate the maximum possible acceleration for which the upper block will not slide.

Part (b): The free body diagram for the given equations is:

Part (c): Calculated value for $n$and$a_x$from the two given equations are$n=14700N$and $a_x=7.84m/s^{-2}$.