Chapter 9, Problem 22P

Part (a)

To determine

The maximum weight of a person who can walk to the extreme end  D  without tipping the beam.

Solution:

Maximum weight of person= $\text{650}\text{N}$

Explanation of Solution

Given:

Weight of the beam, $w=650\text{}\text{N}$

Length of the beam is 20.0 m.

Formula used:

Equilibrium condition: the net force and torque would be zero.

  $\begin{array}{l}{\displaystyle \sum F=0}\\ {\displaystyle \sum \tau =0}\end{array}$

Calculation:

For not tipping, normal force at A is zero. Take the moment about point B.

  $\begin{array}{l}{\displaystyle \sum \tau =0}\\ 650(5)-mg(5)=0\\ mg=650\text{N}\end{array}$

Conclusion:

Thus, the maximum weight of the person who can walk to the extreme end  D  without tipping the beam is 650N.

Part (b)

To determine

The forces that the walls  A  and  B  exert on the beam when the same person is standing at point  D

Answer to Problem 22P

Solution:

  $F_a=0,F_b=1300\text{N}$

Explanation of Solution

Given:

Weight of the beam, $w=650\text{}\text{N}$

Length of the beam is 20.0 m.

Formula used:

Equilibrium condition: the net force and torque would be zero.

  $\begin{array}{l}{\displaystyle \sum F=0}\\ {\displaystyle \sum \tau =0}\end{array}$

Calculation:

When the person moves to the point D, the beams tends to lose the contact A.

  $\begin{array}{l}{\text{ F}}_A=0\\ {\displaystyle \sum F=0}\\ -650-mg+F_B=0\\ -650-650=-F_B\\ F_B=1300\text{N}\end{array}$

Conclusion:

Thus, the forces that the walls  A  and  B  exert on the beam when the same person is standing at point  D are $F_A=0,F_B=1300\text{N}$ .

Part (c)

To determine

The forces that the walls  A  and  B  exert on the beam when the same person is standing 2.0 m to the right of  B .

Answer to Problem 22P

Solution:

  $F_a=812.5\text{N},F_b=487.5\text{N}$

Explanation of Solution

Given:

Weight of the beam, $w=650\text{}\text{N}$

Length of the beam is 20.0 m.

Formula used:

Equilibrium condition: the net force and torque would be zero.

  $\begin{array}{l}{\displaystyle \sum F=0}\\ {\displaystyle \sum \tau =0}\end{array}$

Calculation:

Taking moment about point B:

  $\begin{array}{l}{\displaystyle \sum \tau =0}\\ 650(5)+650(12-2)-F_A(12)=0\\ F_A(12)=9750\\ F_A=812.5\text{N}\\ {\displaystyle \sum F=0}\\ F_A+F_B-650-650=0\\ 812.5+F_B-1300=0\\ F_B=487.5\text{N}\end{array}$

Conclusion:

Thus, the forces that the walls  A  and  B  exert on the beam when the same person is standing at point  D are $F_A=812.5\text{N},F_B=487.5\text{N}$ .