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Q16P

Expert-verifiedFound in: Page 230

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**Calculate \({F_{\rm{A}}}\) and \({F_{\rm{B}}}\) for the beam shown in Fig. 9–56. The downward forces represent the weights of machinery on the beam. Assume that the beam is uniform and has a mass of 280 kg. **

The value of \({F_{\rm{A}}}\) is 6272 N, and the value of \({F_{\rm{B}}}\) is 6072 N.

In equilibrium, the net force in the x and y directions should be zero, and the torque about any point is zero. **For this problem, first, find the condition for equilibrium along the vertical axis and then the condition for zero torque about the left or right end.**

The mass of the beam is \(m = 280\;{\rm{kg}}\).

The length of the beam is \(L = 10\;{\rm{m}}\).

The first force is \({F_1} = 4300\;{\rm{N}}\) at \({r_1} = 2.0\;{\rm{m}}\) from the left end.

The second force is \({F_2} = 3100\;{\rm{N}}\) at \({r_2} = 6.0\;{\rm{m}}\) from the left end.

The third force is \({F_3} = 2200\;{\rm{N}}\) at \({r_3} = 9.0\;{\rm{m}}\) from the left end.

You can assume that the total mass of the beam is at the middle of the beam, i.e., at \(\frac{L}{2} = 5.0\;{\rm{m}}\) from the left end.

The free-body diagram is shown below.

In equilibrium, the net torque about the left end is zero. Then,

\(\begin{array}{l}\left( {{F_{\rm{B}}} \times 10\;{\rm{m}}} \right) - {F_1}{r_1} - {F_2}{r_2} - {F_3}{r_3} - \left( {mg \times \frac{L}{2}} \right) = 0\\{F_{\rm{B}}} \times 10\;{\rm{m}} = {F_1}{r_1} + {F_2}{r_2} + {F_3}{r_3} + \left( {mg \times \frac{L}{2}} \right)\\{F_{\rm{B}}} \times 10\;{\rm{m}} = \left( {4300\;{\rm{N}} \times 2.0\;{\rm{m}}} \right) + \left( {3100\;{\rm{N}} \times 6.0\;{\rm{m}}} \right) + \left( {2200\;{\rm{N}} \times 9.0\;{\rm{m}}} \right) + \left[ {\left( {280\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times 5.0\;{\rm{m}}} \right]\\{F_{\rm{B}}} = 6072\;{\rm{N}}\end{array}\)

Now, for the equilibrium of the forces in the vertical direction,

\(\begin{array}{c}{F_{\rm{A}}} + {F_{\rm{B}}} - {F_1} - {F_2} - {F_3} - mg = 0\\{F_{\rm{A}}} = {F_1} + {F_2} + {F_3} + mg - {F_{\rm{B}}}\\{F_{\rm{A}}} = 4300\;{\rm{N}} + 3100\;{\rm{N}} + 2200\;{\rm{N}} + \left[ {\left( {280\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)} \right] - 6072\;{\rm{N}}\\{F_{\rm{A}}} = 6272\;{\rm{N}}\end{array}\)

Hence, the value of \({F_{\rm{A}}}\) is 6272 N, and the value of \({F_{\rm{B}}}\) is 6072 N.

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