Chapter 8, Problem 76A

To determine

The force exerted by the two personson the board.

Answer to Problem 76A

The force exerted by the person, near the box, on the board is $62.8\text{N}$ and the force exerted by the other person on the board is $37.65\text{N}$ .

Explanation of Solution

Given:

The mass of the wooden board is $m=4.25\text{kg}$ .

The length of the wooden board is $1.75\text{m}$ .

The mass of the box is $m=6.00\text{kg}$ .

Formula used:

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

The expression for the torque as follows:

  $\tau =Fr\sin \theta$

Here, $F$ is the applied force, $r$ is the radius of the wheel, and $\theta$ is the angle made by anobject with the vertical.

Calculation:

Consider the force exerted by the person A on the board is $F_{\text{A}}$ .

Consider the force exerted by the person B,near the box, on the board is $F_{\text{B}}$ .

Consider the force due to the weight of the box is denoted by $F_{\text{g}}$ .

Consider the force due to the centre of gravity is denoted by $F_{\text{cog}}$ .

Consider a crank is connected to a chained wheel.

The forces acting on the board are shown in the Figure below.

  

For the equilibrium of the forces acting on the board,

  $\begin{array}{l}{F}_{\text{A}}+F_{\text{B}}-F_{\text{g}}-F_{\text{cog}}=0\\ F_{\text{A}}+F_{\text{B}}=F_{\text{g}}+F_{\text{cog}}\\ F_{\text{A}}+F_{\text{B}}=\left(m_{\text{board}}g\right)+\left(m_{\text{box}}g\right)\\ F_{\text{A}}+F_{\text{B}}=\left(m_{\text{board}}+m_{\text{box}}\right)g\\ F_{\text{A}}+F_{\text{B}}=\left(4.25\text{kg}+6.00\text{kg}\right)\times 9.80{\text{m/s}}^2\\ F_{\text{A}}+F_{\text{B}}=100.45\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ F_{\text{A}}+F_{\text{B}}=100.45\text{N}\end{array}$

Consider the torque exerted on the board due to the forces $F_{\text{A}},F_{\text{B}},F_{\text{g}},F_{\text{cog}}$ are denoted by

  ${\tau }_{\text{A}},{\tau }_{\text{B}},{\tau }_{\text{g}},{\tau }_{\text{cog}}$ .

For Equilibrium of the torque acting on the board,

  $\begin{array}{l}{\tau }_{\text{A}}+{\tau }_{\text{B}}={\tau }_{\text{g}}+{\tau }_{\text{cog}}\\ 0+{\tau }_{\text{B}}={\tau }_{\text{g}}+{\tau }_{\text{cog}}\\ {\tau }_{\text{B}}={\tau }_{\text{g}}+{\tau }_{\text{cog}}\\ F_{\text{B}}{r}_{\text{B}}=\left[F_{\text{g}}{r}_{\text{g}}\right]+\left[F_{\text{cog}}{r}_{\text{cog}}\right]\\ F_{\text{B}}\left(1.25\text{m}+0.50\text{m}\right)=\left[41.65\text{N}\times 1.25\text{m}\right]+\left[41.65\text{N}\times 1.25\text{m}\right]\\ F_{\text{B}}\left(1.75\text{m}\right)=\left[36.44\text{N}\cdot \text{m}\right]+\left[73.5\text{N}\cdot \text{m}\right]\\ F_{\text{B}}=\left(\frac{109.94\text{N}\cdot \text{m}}{1.75\text{m}}\right)\\ F_{\text{B}}=62.8\text{N}\end{array}$

The force exerted by the person A on the board is

  $\begin{array}{l}{F}_{\text{A}}+F_{\text{B}}=100.45\text{N}\\ F_{\text{A}}+62.8\text{N}=100.45\text{N}\\ {\text{F}}_{\text{A}}=37.65\text{N}\end{array}$

Conclusion:

Therefore, the force exerted by the person, near the box, on the board is $62.8\text{N}$ and the force exerted by the other person on the board is $37.65\text{N}$ .