Question

(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?

Short answer

The vertical support forces at the left and right ends are \(8.0 \times {10^3}\;{\rm{N}}\) and \(5.8 \times {10^3}\;{\rm{N,}}\) respectively.

Step-by-step solution

Meaning of mechanical equilibrium

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 information

Given data:

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

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

Evaluation of vertical support force at the right end of the beam

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.

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

Evaluation of vertical support force at the left end of the beam

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.