Chapter 9, Problem 68GP

(a)

To determine

To Find: The maximum weight of a person.

Answer to Problem 68GP

The maximum weight of a person who can walk to the point D without tipping the beam is $550\text{ N}$ .

Explanation of Solution

Given:

Length of the beam, $l=20\text{ m}$ .

Weight of beam, $m_B\times \text{g }=550\text{ N}$ .

Formula used:

The maximum weight of a person who can walk to the point D without tipping the beam is calculate by applying torque at point B

  $\begin{array}{l}\sum \tau =0\\ m_B\times g\times \text{distance up to B - }W\times \text{distance up to B = 0}\end{array}$

Here,

  $\tau$ is torque,

  $m_B\times g$ is weight of beam,

  $W$ is weight of person.

Calculation:

Substitute the given values in the above equation:

  $\begin{array}{l}{m}_B\times g\times 5-W\times 5=0\\ W={\text{ m}}_B\times g\\ W=550\text{ N}\end{array}$

Conclusion:

Thus, the maximum weight of a person who can walk to the point D without tipping the beam is $550\text{ N}$

(b)

To determine

To Find: The forces exerted by wall A and B on beam.

Answer to Problem 68GP

The forces that the walls A and B exert on the beam when the person is standing at point D is $F_A=0\text{ N and }{F}_B=1100\text{ N}$ .

Explanation of Solution

Given:

Length of the beam, $l=20\text{ m}$ .

Weight of beam, ${\text{m}}_B\times \text{g }=550\text{ N}$ .

Formula used:

At point D, taking the equilibrium condition in vertical direction

  $\begin{array}{l}\sum F_y=0\\ \sum F_y=F_B-m_B\times g-W=0\end{array}$

Here,

  $F_y$ is vertical force,

  $F_B$ is reaction force at B point,

  $m_B\times g$ is weight of beam,

  $W$ is weight of person.

Calculation:

Substitute the given values in the above equation

  $\begin{array}{l}{F}_B-m_B\times g-W=0\\ F_B=m_B\times g+W\\ F_B=2\times 550=1100\text{ N}\end{array}$

Conclusion:

Thus, the forces that the walls A and B exert on the beam when the person is standing at point D is $F_A=0\text{ N and }{F}_B=1100\text{ N}$

(c)

To determine

To Find: The forces exerted by wall A and B on beam.

Answer to Problem 68GP

The forces that the walls A and B exert on the beam when the person is standing at a point 2 m to the right of B is $F_A=137.5\text{ N and }{F}_B=962.5\text{ N}$ .

Explanation of Solution

Given:

Length of the beam, $l=20\text{ m}$ .

Weight of beam, ${\text{m}}_B\times \text{g }=550\text{ N}$ .

Formula used:

Applying the torque at a point $2\text{m}$ to the right of B and then take the equilibrium condition for the vertical forces, Use the following formula

  $\begin{array}{l}\sum {\tau }_B=0\\ m_B\times g\times \text{distance up to B - }W\times \text{distance to the right of B - }{F}_A\times \text{distance uo to B = 0}\\ \sum F_y=0\\ {\text{F}}_{\text{A}}{\text{ + F}}_{\text{B}}{\text{ - m}}_{\text{B}}\times \text{g - W = 0}\end{array}$

Here,

  ${\tau }_B$ is torque at point B,

  $F_y$ is vertical force,

  $F_B$ is reaction force at B point,

  $F_A$ is reaction force at A point,

  $m_B\times g$ is weight of beam,

  $W$ is weight of person.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{F}_A=\frac{m_B\times g\times 5-2\times W}{12}=137.5\text{ N}\\ F_A+F_B-m_B\times g-W=0\\ F_B=2\times W-F_A\\ F_B=1100-137.5=962.5\text{ N}\end{array}$

Conclusion:

Thus, the forces that the walls A and B exert on the beam when the person is standing at a point $2\text{m}$ to the right of B is $F_A=137.5\text{ N and }{F}_B=962.5\text{ N}$ .

(d)

To determine

To Find: The forces exerted by wall A and B on beam.

Answer to Problem 68GP

The forces that the walls A and B exert on the beam when the person is standing at a point $2\text{m}$ to the right of B is $F_A=687.5\text{ N and }{F}_B=412.5\text{ N}$ .

Explanation of Solution

Given:

Length of the beam $l=20\text{ m}$ .

Weight of beam ${\text{m}}_B\times \text{g }=550\text{ N}$ .

Formula used:

Applying the torque at a point $2\text{m}$ to the right of B and then take the equilibrium condition for the vertical forces, following formula will be used

  $\begin{array}{l}\sum {\tau }_A=0\\ m_B\times g\times \text{distance up to B - }W\times \text{distance to the right of A - }{F}_A\times \text{distance uo to B = 0}\\ \sum F_y=0\\ {\text{F}}_{\text{A}}{\text{ + F}}_{\text{B}}{\text{ - m}}_{\text{B}}\times \text{g - W = 0}\end{array}$Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{F}_A=\frac{m_B\times g\times 5+10\times W}{12}=687.5\text{ N}\\ F_A+F_B-m_B\times g-W=0\\ F_B=2\times W-F_A\\ F_B=1100-687.5=412.5\text{ N}\end{array}$

Conclusion:

Thus, the forces that the walls A and B exert on the beam when the person is standing at a point $2\text{m}$ to the right of B is $F_A=687.5\text{ N and }{F}_B=412.5\text{ N}$ .