A 20.0-m-long uniform beam weighing 650 N rests on walls A and B, as shown in Fig. 9–62. (a) Find the maximum weight of a person who can walk to the extreme end D without tipping the beam. Find the forces that the walls A and B exert on the beam when the person is standing: (b) at D; (c) 2.0 m to the right of A.

When a pushing forceFis acting at some point with position vectorr, the resulting torque is the product of the position vector, the force, and the angle between them.

Given data:

The length of the beam is\(L = 20.0\;{\rm{m}}\).

The weight of the beam is \({W_{\rm{B}}} = {m_{\rm{B}}}g = 650\;{\rm{N}}\).

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

Here,\({F_{\rm{A}}}\)is the vertical support force at supportA,\(W\)is the weight of the person, and \({F_{\rm{B}}}\) is the vertical support force at B.

Now, take the torque about support B to calculate the maximum weight of a person who can walk to the extreme end D without tipping the beam.

\(\begin{array}{c}\sum \tau = 0\\{m_{\rm{B}}}g\left( {5.0\;{\rm{m}}} \right) - W\left( {5.0\;{\rm{m}}} \right) = 0\\\left( {650\;{\rm{N}}} \right)\left( {5.0\;{\rm{m}}} \right) - W\left( {5.0\;{\rm{m}}} \right) = 0\\W = 650\;{\rm{N}}\end{array}\)

Thus, the maximum weight of a person who can walk to the extreme end D without tipping the beam is \(650\;{\rm{N}}\).

Since the person is standing at D, the normal force exerted by wall A should be \({F_{\rm{A}}} = 0\).

So, wall B should support the total weight of the beam and the total weight of the person at D as well. Thus, the normal force exerted by wall B can be calculated as:

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{A}}} + {F_{\rm{B}}} - {m_{\rm{B}}}g - W = 0\\\left( 0 \right) + {F_{\rm{B}}} - \left( {650\;{\rm{N}}} \right) - \left( {650\;{\rm{N}}} \right) = 0\\{F_{\rm{B}}} = 1300\;{\rm{N}}\end{array}\)

Thus, the normal forces exerted by wall A and wall B when the person is standing at D are \(0\) and \(1300\;{\rm{N}}\), respectively.

Draw a free-body diagram of the person standing at 2.0 m from wall A toward the right.

Apply the equilibrium of torque about support B.

\(\begin{array}{c}\sum \tau = 0\\{m_{\rm{B}}}g\left( {5.0\;{\rm{m}}} \right) + W\left( {10.0\;{\rm{m}}} \right) - {F_{\rm{A}}}\left( {12.0\;{\rm{m}}} \right) = 0\\\left( {650\;{\rm{N}}} \right)\left( {5.0\;{\rm{m}}} \right) + \left( {650\;{\rm{N}}} \right)\left( {10.0\;{\rm{m}}} \right) - {F_{\rm{A}}}\left( {12.0\;{\rm{m}}} \right) = 0\\{F_{\rm{A}}} = 810\;{\rm{N}}\end{array}\)

Apply the force equilibrium condition along the vertical direction.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{A}}} + {F_{\rm{B}}} - {m_{\rm{B}}}g - W = 0\\\left( {810\;{\rm{N}}} \right) + {F_{\rm{B}}} - \left( {650\;{\rm{N}}} \right) - \left( {650\;{\rm{N}}} \right) = 0\\{F_{\rm{B}}} = 490\;{\rm{N}}\end{array}\)

Thus, the normal forces exerted by wall A and wall B when the person is standing at2.0 m from wall A toward the right are \(810\;{\rm{N}}\) and \(490\;{\rm{N}}\), respectively.