9.2. Solved Examples: Torque; Rotational Equilibrium
Example 1
A [latex]3.4 \text{ m}[/latex] beam is subjected to 2 forces as shown in the figure. F1 is acting at [latex]1.1 \text{ m}[/latex] from the left end and F2 is acting at [latex]1.5 \text{ m}[/latex] from the right end. Calculate the net torque with respect to point A (left end of the beam).
Solution:
The net torque will be calculated as the vector sum of 2 torques: [latex]\tau_1[/latex] created by [latex]F_1[/latex] and [latex]\tau_2[/latex] created by [latex]F_2[/latex]
- The torque of [latex]F_1[/latex] will be positive since it tends to rotate the beam [latex]\text{CCW}[/latex]:
[latex]\tau_1 = F_1 \times r_1 \times \sin\theta_1 = 35 \times 1.1 \times \sin 90^\circ = 38.5 \, \text{Nm}[/latex]
- The torque of [latex]F_2[/latex] will be negative since it tends to rotate [latex]\text{CW}[/latex]:
[latex]\tau_2 = -F_2 \times r_2 \times \sin\theta_2 = -57 \times 1.9 \times \sin 34^\circ = -60.6 \, \text{Nm}[/latex]
Answer:
So, the net torque with respect to A is:
[latex]\tau_{\text{net}} = 38.5 - 60.6 = -22.1 \, \text{Nm}[/latex]
The interpretation of the result is that the beam will rotate [latex]\text{CW}[/latex], since the net torque is negative.
Example 2
A worker of mass [latex]72.0 \text{ kg}[/latex] is standing [latex]1.20 \text{ m}[/latex] away from the left end of a [latex]4.00 \text{ m}[/latex] long beam, which has a mass of [latex]21.0 \text{ kg}[/latex]. What are the weights supported by each end of the beam?
Weights (loads):
The weight of the worker is:
[latex]F_{g1} = m_1 \times g = 72.0 \times 9.8 = 705.6 \, \text{N}[/latex]
The weight of the beam is:
[latex]F_{g2} = m_2 \times g = 21.0 \times 9.8 = 205.8 \, \text{N}[/latex]
Solution:
The system is in linear and rotational equilibrium. This means that the net force and the net torque will be zero.When addressing equilibrium questions, we can express the situation mathematically in two ways:
- Net force and net torque = Zero
[latex]F_{\text{net}} = 0, \quad \tau_{\text{net}} = 0[/latex] (1)
- Sum of all forces pulling up = Sum of all forces pulling down and Sum of all torques rotating [latex]\text{CW}[/latex] = Sum of all torques rotating [latex]\text{CCW}[/latex]
[latex]\sum F_{\text{up}} = \sum F_{\text{down}}[/latex] (2)
– called the first condition of equilibrium (linear equilibrium)
[latex]\sum \tau_{\text{CW}} = \sum \tau_{\text{CCW}}[/latex] (3)
– called the second condition of equilibrium (rotational equilibrium)
To write the two conditions of equilibrium we need to organize the forces and torques in a table of values, as follows:
Point of rotation: A
| [latex]F_{\text{up}}[/latex] | [latex]F_{\text{down}}[/latex] | [latex]\tau_{\text{CW}}[/latex] | [latex]\tau_{\text{CCW}}[/latex] | |
|---|---|---|---|---|
| [latex]F_{\text{g1}}[/latex] | – | 705.6 | 705.6 x 1.20 = 846.72 | – |
| [latex]F_{\text{g2}}[/latex] | – | 205.8 | 205.8 x 2.00 = 411.6 | – |
| [latex]F_1[/latex] | [latex]F_1[/latex] | – | – | – |
| [latex]F_2[/latex] | [latex]F_2[/latex] | – | – | [latex]F_2[/latex] x 4.00 |
| Total | [latex]F_1[/latex] + [latex]F_2[/latex] | 911.4 | 1258.32 | [latex]F_2[/latex] x 4.00 |
Conditions of equilibrium:
[latex]\sum F_{\text{up}} = \sum F_{\text{down}}[/latex] (1)
[latex]F_1 + F_2 = 911.4[/latex]
[latex]\sum \tau_{\text{CW}} = \sum \tau_{\text{CCW}}[/latex] (2)
[latex]F_2 \times 4.00 = 1258.32[/latex]
Solving the system of two equations, results in the following values for the weights supported by each end:
Answer:
[latex]F_1 = 597 \, \text{N}[/latex]
[latex]F_2 = 315 \, \text{N}[/latex]
