Step 1

We are given that the mass of crate is ${m_C} = 50\;{\rm{kg}}$, the mass of dolly is ${m_D} = 10\;{\rm{kg}}$, the coefficient of static friction between casters and floor is ${\mu _f} = 0.35$ and the coefficient of friction between dolly and crate is ${\mu _d} = 0.5$.

We are asked to calculate the maximum force that should be applied without motion of crate.

 
Step 2

The free body diagram of the crate is shown as:

Here, $P$ is the force acting on crate, ${N_d}$ is the reaction force acting on crate, ${F_d}$ is the frictional force acting on crate, ${W_C}$ is the weight of crate and x is the distance between center of gravity and point O.

We have the perpendicular distance of force applied which is ${h_1} = 0.8\;{\rm{m}}$.

We have the total height of crate which is ${h_2} = 1.5\;{\rm{m}}$.

 
Step 3

According to the free body diagram of the crate, the net forces acting on the crate along the y-axis is given as:

\[\begin{array}{c} \sum {F_y} = 0\\ {N_d} - {W_C} = 0\\ {N_d} = {W_C} \end{array}\]…… (1)
 
Step 4

To calculate the weight of crate we use the formula:

\[{W_C} = {m_C}g\]
 
Step 5

Substitute the known value in the equation (1):

\[{N_d} = {m_C}g\]
 
Step 6

Substitute the known value in the equation:

\[\begin{array}{c} {N_d} = \left( {{\rm{60}}\;{\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.81}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ = 588.6\;{\rm{N}} \end{array}\]
 
Step 7

According to the free body diagram of the crate, the net forces acting on the crate along the x-axis is given as:

\[\begin{array}{c} \sum {F_x} = 0\\ P - {F_d} = 0\\ P = {F_d} \end{array}\]…… (2)
 
Step 8

Applying the moment of force equation about point O of the crate:

\[{W_C}x - P{h_1} = 0\]
 
Step 9

Substitute the known value in the equation:

\[\begin{array}{c} \left( {588.6\;{\rm{N}}} \right)x - P\left( {0.8\;{\rm{m}}} \right) = 0\\ P = \frac{{\left( {588.6\;{\rm{N}}} \right)x}}{{\left( {0.8\;{\rm{m}}} \right)}}\\ = \left( {735.75\;{\rm{N/m}}} \right)x \end{array}\]…… (3)
 
Step 10

The free body diagram of the system is shown as:

Here, ${N_B}$ is the reaction force acting on rear wheels, ${N_A}$ is the reaction force acting on front wheels, ${F_A}$ is the frictional force acting on front wheels and ${W_D}$ is the weight of dolly.

We have the perpendicular distance between ground and dolly which is ${h_3} = 0.25\;{\rm{m}}$.

We have the total distance between rear wheels and front wheels which is ${d_1} = 1.5\;{\rm{m}}$.

We have the distance between crate and front wheels which is ${d_2} = 0.25\;{\rm{m}}$.

We have the width of crate which is ${d_3} = 0.6\;{\rm{m}}$.

 
Step 11

According to the free body diagram of the crate, the net forces acting on the system along the y-axis is given as:

\[\begin{array}{c} \sum {F_y} = 0\\ {N_A} + {N_B} - {W_C} - {W_D} = 0\\ {N_A} = {W_C} + {W_D} - {N_B} \end{array}\]…… (4)
 
Step 12

To calculate the weight of dolly we use the formula:

\[{W_D} = {m_D}g\]
 
Step 13

Substitute the known value in the equation (4):

\[{N_A} = {m_C}g + {m_D}g - {N_B}\]
 
Step 14

Substitute the known value in the equation:

\[\begin{array}{c} {N_A} = \left\{ \begin{array}{l} \left( {{\rm{60}}\;{\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.81}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right) + \\ \left( {{\rm{10}}\;{\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.81}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right) - {N_B} \end{array} \right\}\\ = 588.6\;{\rm{N}} + {\rm{98}}{\rm{.1}}\;{\rm{N}} - {N_B}\\ = 686.7\;{\rm{N}} - {N_B} \end{array}\]…… (5)
 
Step 15

According to the free body diagram of the system, the net forces acting on the system along the x-axis is given as:

\[\begin{array}{c} \sum {F_x} = 0\\ P - {F_A} = 0\\ P = {F_A} \end{array}\]…… (6)
 
Step 16

Applying the moment of force equation about point B of the system:

\[{N_A}{d_1} - {W_D}\left( {\frac{{{d_1}}}{2}} \right) - {W_C}\left( {{d_1} - {d_2} - \frac{{{d_3}}}{2}} \right) - P\left( {{h_1} + {h_3}} \right) = 0\]
 
Step 17

Substitute the known value in the equation:

\[{N_A}{d_1} - {m_D}g\left( {\frac{{{d_1}}}{2}} \right) - {m_C}g\left( {{d_1} - {d_2} - \frac{{{d_3}}}{2}} \right) - P\left( {{h_1} + {h_3}} \right) = 0\]
 
Step 18

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} {N_A}\left( {1.5\;{\rm{m}}} \right) - \left( {{\rm{10}}\;{\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.81}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\left( {\frac{{1.5\;{\rm{m}}}}{2}} \right) - \\ \left( {{\rm{60}}\;{\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.81}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\left( {1.5\;{\rm{m}} - 0.25\;{\rm{m}} - \frac{{0.6\;{\rm{m}}}}{2}} \right) - \\ P\left( {0.8\;{\rm{m}} + 0.25\;{\rm{m}}} \right) \end{array} \right\} = 0\\ {N_A}\left( {1.5\;{\rm{m}}} \right) - P\left( {1.05\;{\rm{m}}} \right) = \left\{ \begin{array}{l} 73.58\;{\rm{N}} \cdot {\rm{m}} + \\ 559.17\;{\rm{N}} \cdot {\rm{m}} \end{array} \right\}\\ = 632.75\;{\rm{N}} \cdot {\rm{m}} \end{array}\]…… (7)
 
Step 19

When the crate slips on dolly then to calculate the frictional force acting on dolly, we use the formula:

\[{F_d} = {\mu _d}{N_d}\]
 
Step 20

Substitute the known values in the formula:

\[\begin{array}{c} {F_d} = \left( {0.5} \right)\left( {588.6\;{\rm{N}}} \right)\\ = 294.3\;{\rm{N}} \end{array}\]
 
Step 21

Substitute the known value in equation (2):

\[P = 294.3\;{\rm{N}}\]
 
Step 22

Substitute the known value in equation (3):

\[\begin{array}{c} 294.3\;{\rm{N}} = \left( {{\rm{735}}{\rm{.75}}\;{\rm{N/m}}} \right)x\\ x = \frac{{\left( {294.3\;{\rm{N}}} \right)}}{{\left( {{\rm{735}}{\rm{.75}}\;{\rm{N/m}}} \right)}}\\ = 0.4\;{\rm{m}} \end{array}\]

Since the width of crate is $0.6\;{\rm{m}}$ and the value of x is greater than the half of width of crate therefore, the crate tips on dolly and now we have to calculate the force at $x = 0.3\,{\rm{m}}$.

 
Step 23

Substitute the known value in equation (3):

\[\begin{array}{c} P = \left( {{\rm{735}}{\rm{.75}}\;{\rm{N/m}}} \right)\left( {{\rm{0}}{\rm{.3}}\;{\rm{m}}} \right)\\ = 220.725\;{\rm{N}} \end{array}\]
 
Step 24

Substitute the known value in equation (2):

\[{F_d} = 220.725\;{\rm{N}}\]
 
Step 25

Now, when the dolly slips at front wheels then to calculate the frictional force acting on A, we use the formula:

\[{F_A} = {\mu _f}{N_A}\]
 
Step 26

Substitute the known value from equation (6):

\[\begin{array}{c} P = {\mu _f}{N_A}\\ {N_A} = \frac{P}{{{\mu _f}}} \end{array}\]
 
Step 27

Substitute the known value in equation (7):

\[\begin{array}{c} \left( {\frac{P}{{0.35}}} \right)\left( {1.5\,{\rm{m}}} \right) - P\left( {1.05\;{\rm{m}}} \right) = 632.75\;{\rm{N}} \cdot {\rm{m}}\\ P\left( {4.28\;{\rm{m}} - 1.05\;{\rm{m}}} \right) = 632.75\;{\rm{N}} \cdot {\rm{m}}\\ P = \frac{{\left( {632.75\;{\rm{N}} \cdot {\rm{m}}} \right)}}{{\left( {3.23\;{\rm{m}}} \right)}}\\ = 195.9\;{\rm{N}} \end{array}\]