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Physics Principles with Applications
Found in: Page 230
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

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.

See the step by step solution

Step by Step Solution

Step 1: Concepts

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.

Step 2: Given data

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.

Step 3: Calculation

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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