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Q10P

Expert-verifiedFound in: Page 230

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

The forces on A and B are \({F_{\rm{A}}} = - 2928\;{\rm{N}}\) and \({F_{\rm{B}}} = 14700\;{\rm{N}}\) , respectively.

The force on point A is \({F_{\rm{A}}}\).

The force on point B is \({F_{\rm{B}}}\).

The mass of the cantilever is \(m = 1200\;{\rm{kg}}\).

**In order to determine the forces at points A and B, first, find the torque about the left end of the beam and apply the conditions of equilibrium.**

The following is the free body diagram.

The relation to calculate the net torque can be written as:

** **

\(\begin{array}{c}\sum \tau = 0\\\left( {{F_{\rm{B}}} \times 20\;{\rm{m}}} \right) - \left( {mg \times 25\;{\rm{m}}} \right) = 0\end{array}\)** **

Here, \(g\) is the gravitational acceleration and \({F_{\rm{B}}}\) is the force on point B.* *

On plugging the values in the above relation, you get:

\(\begin{array}{c}\left( {{F_{\rm{B}}} \times 20\;{\rm{m}}} \right) - \left( {1200\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times 25\;{\rm{m}}} \right) = 0\\{F_{\rm{B}}} = 14700\;{\rm{N}}\end{array}\)

The relation to calculate the force on the beam can be written as:

** **

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{A}}} + {F_{\rm{B}}} - mg = 0\end{array}\)** **

Here, \({F_{\rm{A}}}\) is the force on point A.

** **

On plugging the values in the above relation, you get:

\(\begin{array}{c}{F_{\rm{A}}} + \left( {14700\;{\rm{N}}} \right) - \left( {1200\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) = 0\\{F_{\rm{A}}} = - 2928\;{\rm{N}}\end{array}\)

Thus, the forces on points A and B are \({F_{\rm{A}}} = - 2928\;{\rm{N}}\) and \({F_{\rm{B}}} = 14700\;{\rm{N}}\), respectively.

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