Chapter 8.3, Problem 41PP

To determine

The forces on the two supports

Answer to Problem 41PP

The force on support at one endis $-8.3\times {10}^2\text{N}$ .

The force on supportat the middle is $1.8\times {10}^3\text{N}$ .

Explanation of Solution

Given:

The mass of the diver is $m_{\text{diver}}=85\text{kg}$ .

The length of the board is $l=3.5\text{m}$ .

The mass of the board is $m_{\text{board}}=14\text{kg}$ .

Formula used:

The torque is the product of the lever arm and the applied force.

The expression for the torque is,

  $\tau =Fr$

Here, $F$ is the applied force, $r$ is the lever arm.

Calculation:

The forces acting on the board is shown in the figure below.

  

Consider the weight of the board is concentrated at the centre support B.

The axis of rotation passes through the support B.

Consider the self weight of the diver is denoted by $F_{\text{diver}}$ .

The torque produce due to the self weight of the board $\left(F_{\text{board}}\right)$ and the torque produced by the force due to support B$\left(F_B\right)$ are zero as both the centre of mass of the board and the axis of rotation passes through the support B.

Consider the torque produced by the support $A$ and torque produced due the self weight of the diver is denoted by ${\tau }_{\text{A}}$ and ${\tau }_{\text{diver}}$ .

For the rotational equilibrium of the board,

  $\begin{array}{l}{\tau }_{\text{A}}+{\tau }_{\text{diver}}=0\\ F_{\text{A}}{r}_{\text{A}}+F_{\text{diver}}{r}_{\text{diver}}=0\\ F_{\text{A}}=-\left(\frac{F_{\text{diver}}{r}_{\text{diver}}}{r_{\text{A}}}\right)\\ F_{\text{A}}=-\left(\frac{m_{\text{diver}}g\times r_{\text{diver}}}{r_{\text{A}}}\right)\\ F_{\text{A}}=-\left(\frac{85\text{kg}\times 9.80{\text{m/s}}^2\times 1.75\text{m}}{1.75\text{m}}\right)\\ F_{\text{A}}=-8.3\times {10}^2\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{\text{1}\text{kg}\cdot {\text{m/s}}^2}\right)\\ F_{\text{A}}=-8.3\times {10}^2\text{N}\end{array}$

The board is not moving and the net vertical forces acting on the board is zero.

For the vertical equilibrium of forces,

  $F_{\text{A}}+F_{\text{B}}=F_{\text{board}}+F_{\text{diver}}$

The value of the torque ${\tau }_{\text{A}}=-{\tau }_{\text{diver}}$ . Hence, the value of corresponding forces $F_{\text{A}}=-F_{\text{diver}}$ .

  $\begin{array}{l}-F_{\text{diver}}+F_{\text{B}}=F_{\text{board}}+F_{\text{diver}}\\ F_{\text{B}}=F_{\text{board}}+2F_{\text{diver}}\\ F_{\text{B}}=m_{\text{board}}g+2m_{\text{diver}}g\\ F_{\text{B}}=\left(m_{\text{board}}+2m_{\text{diver}}\right)g\end{array}$

  $\begin{array}{l}{F}_{\text{B}}=\left(14\text{kg}+2\times 85\text{kg}\right)\times 9.80{\text{m/s}}^2\\ F_{\text{B}}=1.8\times {10}^3\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ F_{\text{B}}=1.8\times {10}^3\text{N}\end{array}$

Conclusion:

Therefore, the force exerted on support $A$ is $-8.3\times {10}^2\text{N}$ and the force exerted on support $B$ is $1.8\times {10}^3\text{N}$ .