Step 1

We have given the following values:

The mass of the crate is $M = 150\;{\rm{kg}}$.

The mass of the cart is $m = 10\;{\rm{kg}}$.

The coefficient of static friction between the crate and the cart is ${\mu _s} = 0.2$.

We are asked to calculate the maximum force P that can be applied to the handle without causing the crate to tip or slip on the cart.

 
Step 2

Assume that the crate slips before it tips.

Draw a free-body diagram and a kinetic diagram of the crate.

Here, G is the center of gravity, g is the acceleration due to gravity, f is the friction at point A, N is the normal force at point A, and a is the acceleration of the crate.

 
Step 3

Consider figure (a) and apply equilibrium equation of motion of crate along the y-axis:

\[\begin{array}{c} \sum {{F_y}} = M{a_y}\\ N - Mg = M\left( 0 \right)\\ N = Mg \end{array}\]

Substitute the value of M and g in the above equation:

\[\begin{array}{c} N = \left( {150\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m}}/{{\rm{s}}^2}} \right) \times \left( {\frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}}}} \right)\\ N = 1471.5\;{\rm{N}} \end{array}\]
 
Step 4

Consider figure (a) and apply equilibrium equation of motion of crate along the x-axis:

\[\begin{array}{c} \sum {{F_x}} = M{a_x}\\ f = Ma\\ {\mu _s}N = Ma\\ a = \frac{{{\mu _s}N}}{M} \end{array}\]

Substitute the value of N, M, and ${\mu _s}$ in the above equation:

\[\begin{array}{c} a = \frac{{\left( {0.2} \right)\left( {1471.5\;{\rm{N}}} \right) \times \left( {\frac{{1\;{\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}}}{{1\;{\rm{N}}}}} \right)}}{{150\;{\rm{kg}}}}\\ = 1.962\;{\rm{m}}/{{\rm{s}}^2} \end{array}\]
 
Step 5

Consider figure (a) and apply the equilibrium equation for the moment about point A:

\[\begin{array}{c} \sum {{M_A}} = \sum {{{\left( {{M_k}} \right)}_A}} \\ Mg \times x = Ma \times 0.5\;{\rm{m}}\\ g \times x = a \times 0.5\;{\rm{m}}\\ x = \frac{{a \times 0.5\;{\rm{m}}}}{g} \end{array}\]

Substitute the value of a and g in the above equation:

\[\begin{array}{c} x = \frac{{\left( {1.962\;{\rm{m}}/{{\rm{s}}^2}} \right) \times 0.5\;{\rm{m}}}}{{\left( {9.81\;{\rm{m}}/{{\rm{s}}^2}} \right)}}\\ x = 0.1\;{\rm{m}} \end{array}\]

Since $x = 0.1\;{\rm{m}} < 0.25\;{\rm{m}}$, the crate indeed slips before it tips.

 
Step 6

Draw a free-body diagram of the complete system.

Here, ${N_1}$ and ${N_2}$ are the normal reaction forces.

 
Step 7

Consider figure (b) and apply the equilibrium equation of motion along the horizontal direction of motion:

\[\begin{array}{c} \sum {{F_x}} = \left( {m + M} \right)a\\ P = \left( {m + M} \right)a \end{array}\]

Substitute the value of m, M, and a in the above equation:

\[\begin{array}{c} P = \left( {10\;{\rm{kg}} + 150\;{\rm{kg}}} \right)\left( {1.962\;{\rm{m}}/{{\rm{s}}^2}} \right)\\ = \left( {313.92\;{\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}} \right) \times \left( {\frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}}}} \right)\\ = 313.92\;{\rm{N}}\\ \approx 314\;{\rm{N}} \end{array}\]