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An 84 kg adult sits at one end of an 8.9 m long board, while his 22 kg child sits on the other end. What is the total torque exerted on the board?

Answer :

The total torque exerted on the board by the adult and child is 4659.92 Nm.

To calculate the total torque exerted on the board, we need to find the torque exerted by each person and then add them together.

The torque exerted by a force about a pivot point is given by the formula:

[tex]\[ \tau = r \times F \][/tex]

Where:

  • [tex]\(\tau\)[/tex] is the torque,
  • [tex]\(r\)[/tex] is the distance from the pivot point to the point where the force is applied
  • [tex]\(F\)[/tex]is the magnitude of the force.

For the adult:

  • Mass [tex]\(m_1 = 84 \, \text{kg}\)[/tex]
  • Distance from the pivot[tex]\(r_1 = \frac{8.9}{2} \, \text{m} = 4.45 \, \text{m}\)[/tex]
  • Gravitational acceleration [tex]\(g = 9.81 \, \text{m/s}^2\)[/tex]

The torque exerted by the adult is:

[tex]\[ \tau_1 = r_1 \times (m_1 \times g) = 4.45 \, \text{m} \times (84 \, \text{kg} \times 9.81 \, \text{m/s}^2) \][/tex]

For the child:

  • Mass [tex]\(m_2 = 22 \, \text{kg}\)[/tex]
  • Distance from the pivot [tex]\(r_2 = \frac{8.9}{2} \, \text{m} = 4.45 \, \text{m}\)[/tex]

The torque exerted by the child is:

[tex]\[ \tau_2 = r_2 \times (m_2 \times g) = 4.45 \, \text{m} \times (22 \, \text{kg} \times 9.81 \, \text{m/s}^2) \][/tex]

The total torque is the sum of these two torques:

[tex]\[ \tau_{\text{total}} = \tau_1 + \tau_2 \][/tex]

Let's calculate this.

First, let's calculate the torques for the adult and the child:

For the adult:

[tex]\[ \tau_1 = 4.45 \, \text{m} \times (84 \, \text{kg} \times 9.81 \, \text{m/s}^2) \][/tex]

For the child:

[tex]\[ \tau_2 = 4.45 \, \text{m} \times (22 \, \text{kg} \times 9.81 \, \text{m/s}^2) \][/tex]

Now, let's calculate the total torque:

[tex]\[ \tau_{\text{total}} = \tau_1 + \tau_2 \][/tex]

Calculating:

[tex]\[ \tau_1 = 4.45 \, \text{m} \times (84 \, \text{kg} \times 9.81 \, \text{m/s}^2) = 3688.46 \, \text{Nm} \][/tex]

[tex]\[ \tau_2 = 4.45 \, \text{m} \times (22 \, \text{kg} \times 9.81 \, \text{m/s}^2) = 971.46 \, \text{Nm} \]\[ \tau_{\text{total}} = 3688.46 \, \text{Nm} + 971.46 \, \text{Nm} = 4659.92 \, \text{Nm} \][/tex]

So, the total torque exerted on the board is [tex]\( 4659.92 \, \text{Nm} \).[/tex]

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Rewritten by : Barada