(II) A uniform steel beam has a mass of 940 kg. On it is resting half of an identical beam, as shown in Fig. 9–60. What is the vertical support force at each end?

A body is supposed to be in mechanical equilibrium when the addition of all the forces working on the body equals zero and the body's motion does not change.

Given data:

The mass of the beam is\(M = 940\;{\rm{kg}}\).

The length of the beam is \(l\).

The free-body diagram of the beam can be drawn as:

Here,\({F_{\rm{A}}}\)is the force on the beam due to the left support and \({F_{\rm{B}}}\) is the force on the beam due to the right support.

Apply the condition of rotational equilibrium.

\(\begin{array}{c}\sum \tau = 0\\{F_{\rm{B}}}l - \left( {\frac{{Mg}}{2}} \right)\left( {\frac{l}{4}} \right) - \left( {\frac{{Mgl}}{2}} \right) = 0\\{F_{\rm{B}}} = \frac{{Mg}}{8} + \frac{{Mg}}{2}\\{F_{\rm{B}}} = \frac{{5Mg}}{8}\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{l}{F_{\rm{B}}} = \frac{{5\left( {940\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)}}{8}\\{F_{\rm{B}}} = 5.8 \times {10^3}\;{\rm{N}}\end{array}\)

Apply the translational equilibrium condition along the y-direction.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{A}}} + {F_{\rm{B}}} - Mg - \frac{1}{2}Mg = 0\\{F_{\rm{A}}} + {F_{\rm{B}}} - \frac{3}{2}Mg = 0\\{F_{\rm{A}}} = - {F_{\rm{B}}} + \frac{3}{2}Mg\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}{F_{\rm{A}}} = - \left( {5.8 \times {{10}^3}\;{\rm{N}}} \right) + \frac{3}{2}\left( {940\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\\{F_{\rm{A}}} = 8.0 \times {10^3}\;{\rm{N}}\end{array}\)

Thus, the forces on the beam due to the left and right supports are \(8.0 \times {10^3}\;{\rm{N}}\) and \(5.8 \times {10^3}\;{\rm{N}}\), respectively.