Step 1

(a)

We are given the weight of block $W = 200\;{\rm{N}}$, the coefficient of static friction as ${\mu _s} = 0.3$.

We are asked to determine the force required to move the block.

 
Step 2

We will draw a free body diagram of the block.

Here, $N$ is the reaction force and $f = {\mu _s}N$ is the friction force.

The length $AD$ is $AD = 2\;{\rm{m}}$.

The length $AB$ is $AB = 1\;{\rm{m}}$.

The length $AO$ is $AO = 0.5\;{\rm{m}}$.

 
Step 3

Consider the block is slipping on the ground.

We will resolve the forces in the vertical direction.

\[\begin{array}{c} \sum {{F_y}} = 0\\ N - W = 0\\ N = W \end{array}\] …. (1)

Substitute the given value in the above equation.

\[N = 20{\rm{0}}\;{\rm{N}}\]
 
Step 4

We will find the friction force.

\[f = {\mu _s}N\]

Substitute the given value in the above equation.

\[\begin{array}{c} f = 0.3\left( {200\;{\rm{N}}} \right)\\ f = 60\;{\rm{N}} \end{array}\]
 
Step 5

We will take the moment about point $O$.

\[\begin{array}{c} \sum {{M_O}} = 0\\ Nx - \left( {P \times AD} \right) = 0\\ Nx = P \times AD \end{array}\] …. (2)

In the case of impending, the applied force just equals to the frictional force.

So, $f = P$

Substitute the given value in the equation (2).

\[\begin{array}{c} \left( {20{\rm{0}}\;{\rm{N}}} \right)x = \left( {{\rm{60}}\;{\rm{N}}} \right) \times \left( {{\rm{2}}\;{\rm{m}}} \right)\\ x = 0.6 \end{array}\]

The distance between the point $O$ and the normal force is greater than $0.5\;{\rm{m}}$, so the block will tip on the ground.

 
Step 6

We will take the moment about point $O$ to find the required force to cause impending motion.

\[\begin{array}{c} \sum {{M_O}} = 0\\ Nx - \left( {P \times AD} \right) = 0\\ P = \frac{{Nx}}{{AD}} \end{array}\]

Substitute the given value in the above equation to find the required force for $x = 0.5\;{\rm{m}}$.

\[\begin{array}{c} P = \frac{{\left( {{\rm{200}}\;{\rm{N}}} \right)\left( {{\rm{0}}{\rm{.5}}\;{\rm{m}}} \right)}}{{\left( {2\;{\rm{m}}} \right)}}\\ P = 50\;{\rm{N}} \end{array}\]

So, the required force is $P = 50\;{\rm{N}}$.

 
Step 7

(b)

We are given the weight of block $W = 100\;{\rm{N}}$, the coefficient of static friction as ${\mu _s} = 0.4$.

We are asked to determine the force required to move the block.

 
Step 8

We will draw a free body diagram of the block.

Here, $N$ is the reaction force and $f = {\mu _s}N$ is the friction force.

The length $AD$ is $AD = 1\;{\rm{m}}$.

The length $AB$ is $AB = 1\;{\rm{m}}$.

The length $AO$ is $AO = 0.5\;{\rm{m}}$.

 
Step 9

Consider the block is slipping on the ground.

We will resolve the forces in the vertical direction.

\[\begin{array}{c} \sum {{F_y}} = 0\\ N - W = 0\\ N = W \end{array}\] …. (3)

Substitute the given value in the above equation.

\[N = 10{\rm{0}}\;{\rm{N}}\]
 
Step 10

We will find the friction force.

\[f = {\mu _s}N\]

Substitute the given value in the above equation.

\[\begin{array}{c} f = 0.4\left( {00\;{\rm{N}}} \right)\\ f = 40\;{\rm{N}} \end{array}\]
 
Step 11

We will take the moment about point $O$.

\[\begin{array}{c} \sum {{M_O}} = 0\\ Nx - \left( {P \times AD} \right) = 0\\ Nx = P \times AD \end{array}\] …. (4)

In the case of impending, the applied force just equals to the frictional force.

So, $f = P$

Substitute the given value in the equation (2).

\[\begin{array}{c} \left( {10{\rm{0}}\;{\rm{N}}} \right)x = \left( {{\rm{40}}\;{\rm{N}}} \right) \times \left( {1\;{\rm{m}}} \right)\\ x = 0.4 \end{array}\]

The distance between the point $O$ and the normal force is less than $0.5\;{\rm{m}}$, so the block will slip on the ground.

So, the required force is $P = 40\;{\rm{N}}$.