To calculate the tension in the cable, we’ll need to apply the principles of static equilibrium. Since the plank is in equilibrium, the sum of the torques around any point (usually the pivot point) must equal zero. Additionally, the sum of the forces in both the vertical and horizontal directions must also be zero.

Given Data:

• Length of the plank, $L = 2.200 \, \text{m}$
• Mass of the plank, $M = 7.00 \, \text{kg}$
• Height of cable above the pivot, $H = 1.30 \, \text{m}$
• Distance of center of mass from the pivot, $d = 0.600 \, \text{m}$
• Gravitational acceleration, $g = 9.81 \, \text{m/s}^2$

Steps:

1. Calculate the Weight of the Plank:
   $W = Mg = 7.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 68.67 \, \text{N}$

2. Set up Torque Equilibrium: We’ll take the pivot point as our reference for torques. The torques due to all forces acting around this point must add to zero.

   • The weight of the plank acts downward at the center of mass, located $d = 0.600 \, \text{m}$ from the pivot.
   • The tension $T$ in the cable acts at an angle due to its height $H$ above the pivot and length $L$ of the plank.

3. Torque Due to the Weight of the Plank: The torque due to the weight of the plank about the pivot is: $\tau_{\text{weight}} = W \cdot d = 68.67 \, \text{N} \times 0.600 \, \text{m} = 41.202 \, \text{Nm}$

4. Torque Due to the Tension in the Cable: The cable creates a counteracting torque. To find the torque arm (distance from pivot to where the force acts perpendicular to the plank), we calculate the length of the cable section above the pivot, which is a diagonal distance from pivot to where the cable connects: $D = \sqrt{L^2 + H^2} = \sqrt{(2.200)^2 + (1.30)^2} = \sqrt{4.84 + 1.69} = \sqrt{6.53} = 2.555 \, \text{m}$

   The angle $\theta$ between the cable and the plank can be found using: $\sin \theta = \frac{H}{D} = \frac{1.30}{2.555} = 0.509 \implies \theta = \sin^{-1}(0.509) = 30.6^\circ$

   Thus, the torque due to tension $T$ is: $\tau_{\text{tension}} = T \cdot L \cdot \sin(\theta)$

5. Equating Torques for Equilibrium: For rotational equilibrium: $\tau_{\text{tension}} = \tau_{\text{weight}}$ $T \cdot L \cdot \sin(\theta) = W \cdot d$ Solving for $T$: $T = \frac{W \cdot d}{L \cdot \sin(\theta)}$ Substitute the values: $T = \frac{68.67 \cdot 0.600}{2.200 \cdot 0.509} = \frac{41.202}{1.1198} = 36.8 \, \text{N}$

Final Answer:

The tension in the cable is approximately: $T = 36.8 \, \text{N}$

Would you like further clarification on any of these steps, or have any questions?

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Related Questions:

1. How would the tension change if the height $H$ were increased?
2. What would happen to the tension if the mass of the plank doubled?
3. How does the position of the center of mass affect the tension?
4. If the angle of the cable to the plank were changed, how would you calculate the new tension?
5. What are the horizontal and vertical force components at the pivot?

Tip: When solving for torques, choosing the pivot as the point of reference simplifies calculations by eliminating the torque due to forces acting directly on it.