Step 1

We are given the following data:

The weight of a man is ${W_1} = 200\;{\rm{lb}}$.

The weight of a crate is ${W_2} = 450\;{\rm{lb}}$.

The coefficient of static friction between the crate and floor is ${\mu _s} = 0.3$.

The coefficient of static friction between the man shoes and floor is $\mu {'_s} = 0.6$.

The height of the crate is $h = 3\;{\rm{ft}}$.

The width of the crate is $b = 2\;{\rm{ft}}$.

We are asked to determine if the man can move the crate.

 
Step 2

We will draw a free body diagram of the crate.

Here, ${N_C}$ is the normal reaction force at crate and ${F_C}$ represent the frictions force between the floor and crate and ${W_2}$ represent the weight of the crate.

 
Step 3

We will take the sum of the force along the horizontal direction and equate to zero.

\[\begin{array}{c} \sum {{F_x}} = 0\\ \left[ {{F_C} - P} \right] = 0\\ {F_C} = P \end{array}\]
 
Step 4

We will take the sum of the force along the vertical direction and equate to zero.

\[\begin{array}{c} \sum {{F_y}} = 0\\ \left[ {{N_C} - {W_2}} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ {{N_C} - \left( {450\;{\rm{lb}}} \right)} \right] = 0\\ {N_C} = \left( {450} \right)\;{\rm{lb}} \end{array}\]
 
Step 5

We will take the sum of the moment about point $O$.

\[\begin{array}{c} \sum {{M_O}} = 0\\ \left[ {\left( P \right)\left( h \right) - \left( {{N_C}} \right)\left( x \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ {P\left( {3\;{\rm{ft}}} \right) - \left( {450\;{\rm{lb}}} \right)\left( x \right)} \right] = 0\\ \left( {{\rm{3}}\;{\rm{ft}}} \right)P = \left( {450x} \right)\;{\rm{lb}}\\ P = \left( {150x} \right)\;{\rm{lb/ft}} \end{array}\]
 
Step 6

We will draw the free-body diagram of the man.

Here, ${N_m}$ is the normal reaction force at man and ${F_m}$ represent the frictions force between the floor and man shoes and ${W_1}$ represent the weight of the man.

We will take the sum of the force along the horizontal direction and equate to zero.

\[\begin{array}{c} \sum {{F_x}} = 0\\ \left[ {P - {F_m}} \right] = 0\\ {F_m} = P \end{array}\]
 
Step 7

We will take the sum of the force along the vertical direction and equate to zero.

\[\begin{array}{c} \sum {{F_y}} = 0\\ \left[ {{N_m} - {W_1}} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ {{N_m} - \left( {200\;{\rm{lb}}} \right)} \right] = 0\\ {N_m} = \left( {200} \right)\;{\rm{lb}} \end{array}\]
 
Step 8

We will assume that the crate slides before tipping.

The formula to calculate the friction force between the crate and floor ${F_C}$ is given by,

\[{F_C} = {\mu _s}{N_C}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_C} = \left( {0.3} \right)\left( {450\;{\rm{lb}}} \right)\\ = 135\;{\rm{lb}} \end{array}\]
 
Step 9

Substitute the value ${F_C} = 135\;{\rm{lb}}$ in the equation ${F_C} = P$ to obtain $P$.

\[P = 135\;{\rm{lb}}\]
 
Step 10

Substitute the value $P = 135\;{\rm{lb}}$ in the equation ${F_m} = P$ to obtain ${F_m}$.

\[{F_m} = 135\;{\rm{lb}}\]
 
Step 11

Substitute the value $P = 135\;{\rm{lb}}$ in the equation $P = \left( {150x} \right)\;{\rm{lb/ft}}$ to obtain $x$.

\[\begin{array}{c} \left( {135} \right)\;{\rm{lb}} = \left( {150x} \right)\;{\rm{lb/ft}}\\ x = 0.9\;{\rm{ft}} \end{array}\]
 
Step 12

Since, $x < \left( {\frac{b}{2}} \right)$ means $x < \left( {1\;{\rm{ft}}} \right)$, the crate indeed slides before tipping as assumed.

The formula to calculate the maximum friction force between the man shoes and the floor is given by,

\[{\left( {{F_m}} \right)_{{\rm{max}}}} = \left( {\mu {'_s}} \right)\left( {{N_C}} \right)\]

Substitute all the known values in the above formula to obtain ${\left( {{F_m}} \right)_{\max }}$.

\[\begin{array}{c} {\left( {{F_m}} \right)_{{\rm{max}}}} = \left( {0.6} \right)\left( {200\;{\rm{lb}}} \right)\\ = 120\;{\rm{lb}} \end{array}\]

Since, $\left( {{F_m}} \right) > {\left( {{F_m}} \right)_{\max }}$, the man slips. Therefore he is not able to move the crate.