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 $P = 40\;{\rm{N}}$.
