(II) A 2500-kg trailer is attached to a stationary truck at point B, Fig. 9–61. Determine the normal force exerted by the road on the rear tires at A, and the vertical force exerted on the trailer by the support B.

Translational equilibrium is considered the first condition for equilibrium, and rotational equilibrium is considered the second condition of equilibrium.

In the case of translational equilibrium, there is no resultant force acting on the body.

Given data:

The mass of the trailer is \(M = 2500\;{\rm{kg}}\).

The free-body diagram can be drawn as:

Here,\({F_{\rm{A}}}\)is the normal force exerted by the road on the rear tires at A,\({F_{\rm{B}}}\)is the vertical force exerted on the trailer by support B,\({x_1} = 5.5\;{\rm{m}}\)denotes thethe distance between force\({F_{\rm{B}}}\)and force\(Mg\), and\({x_2} = 2.5\;{\rm{m}}\)denotesthe distance between force\({F_{\rm{A}}}\)and\(Mg\).

Now, take the torque about point B to calculate the normal force exerted by the road on the rear tires at A.

\(\begin{array}{c}\sum \tau = 0\\Mg{x_1} - {F_{\rm{A}}}\left( {{x_1} + {x_2}} \right) = 0\\\left( {2500\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {5.5\;{\rm{m}}} \right) - {F_A}\left[ {\left( {5.5\;{\rm{m}}} \right) + \left( {2.5\;{\rm{m}}} \right)} \right] = 0\\{F_{\rm{A}}} = 16844\;{\rm{N}}\end{array}\)

Now, apply the translational equilibrium condition along the vertical direction.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{B}}} + {F_{\rm{A}}} - Mg = 0\\{F_{\rm{B}}} + \left( {16844\;{\rm{N}}} \right) - \left( {2500\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right) = 0\\{F_{\rm{B}}} = 7656\;{\rm{N}}\end{array}\)

Thus, the normal force exerted by the road on the rear tires at A is \(16844\;{\rm{N}}\) , and the vertical force exerted on the trailer by support B is \(7656\;{\rm{N}}\).