Problem 59 - College Physics

Question

Prior to being placed in its hole, a \(5700-\mathrm{N}, 9.0\) -m-long, uniform utility pole makes some nonzero angle with the vertical. A vertical cable attached 2.0 \(\mathrm{m}\) below its upper end holds it in place while its lower end rests on the ground. (a) Find the tension in the cable and the magnitude and direction of the force exerted by the ground on the pole. (b) Why don't we need to know the angle the pole makes with the vertical, as long as it is not zero?

Step-by-Step Solution

Answer

(a) Tension in the cable: 3664 N; ground force: 2036 N vertical. (b) The angle is irrelevant as torques depend on distances, not angle.

Step 1: Identify the Forces

The problem involves a uniform utility pole held by a vertical cable. The forces acting on the pole are gravity (weight of the pole), tension in the cable, and the normal force and friction force from the ground. The weight of the pole acts at its center of gravity, 4.5 m from the base (half of 9.0 m). The tension is acting 2.0 m from the top, or equivalently, 7.0 m from the base.

Step 2: Set Up Equations for Torque

Consider the pole's base as the pivot point to analyze the torques. The net torque around this point must be zero because the pole is in equilibrium. The torque due to the weight is calculated as follows: \[ \tau_{\text{weight}} = 5700\, \text{N} \times 4.5\, \text{m}. \] The torque due to the tension in the cable is: \[ \tau_{\text{tension}} = T \times 7.0\, \text{m}. \] Set the torques equal to solve for the tension (T).

Step 3: Solve for Tension in the Cable

Equating the torques, we have \[ T \times 7.0 = 5700 \times 4.5. \] Solving for \( T \), \[ T = \frac{5700 \times 4.5}{7.0}. \] Calculate to find \( T \approx 3664 \text{ N}. \)

Step 4: Use Equilibrium to Find Forces Exerted by Ground

Now analyze the vertical force balance. Since the pole is not accelerating vertically, the vertical forces must sum to zero. The normal force \( N \) plus the vertical component of the tension should equal the weight of the pole: \[ N = 5700 - T. \] Substitute \( T = 3664 \text{ N} \) into the equation to find \( N. \)

Step 5: Calculate Normal Force

Substitute the calculated tension into the equation for the normal force to solve: \[ N = 5700 - 3664 = 2036 \text{ N}. \] This is the force exerted by the ground vertically.

Step 6: Understand Why Angle is Irrelevant for Torque Calculation

Considering why the angle of the pole with the vertical is unimportant, we note that by choosing the point of contact with the ground as the pivot, the lever arm for torque due to both gravity and tension is defined without needing the angle. They only depend on the distances from the pivot point, which are given directly.

Key Concepts

Torque

Torque is a measure of how much a force causes an object to rotate around an axis or pivot point. In our utility pole example, torque helps us understand how forces like the pole's weight and the cable's tension make the pole rotate about its base.

• Torque depends on the magnitude of the force and the distance from the pivot point, known as the lever arm.
• The direction of the force also affects the torque - it needs to be perpendicular to the lever arm to have maximum effect.
• In mathematical terms, torque \( \tau \) can be expressed as \( \tau = F \times r \times \sin(\theta) \), where \( F \) is the force applied, \( r \) is the lever arm, and \( \theta \) is the angle between the force and lever arm.

In this exercise, since the pole is in equilibrium, we set the sum of torques to zero, balancing the torque from the pole's weight against the torque from the cable's tension.

Equilibrium

Equilibrium occurs when all the forces and torques acting on an object are balanced, leading to no net acceleration. In the case of the utility pole, it remains stationary because it is in equilibrium.

• For equilibrium, both the net force and the net torque must be zero.
• This means that all upward forces must balance the downward forces, and all clockwise torques must balance counterclockwise torques.
• Given the forces involved - tension in the cable and normal force from the ground - precise calculations ensure that these conditions are satisfied.

Understanding equilibrium is crucial in solving this problem, as it helps determine the unknown forces and verify that the system is stable.

Tension

Tension refers to the force exerted by a rope or cable when it holds or supports an object. In this scenario, the cable provides tension 2 meters from the top of the pole, preventing it from falling.

• Tension acts along the length of the cable in equal and opposite directions at both ends.
• In physics problems, tension can be solved by considering the forces and torques around the pivot point, as we did in finding the force of tension as approximately 3664 N.
• The tension keeps the pole from falling, effectively replacing the need for any horizontal support at the top.

Tension is critical in understanding how cables and other supports work to stabilize structures.

Normal Force

Normal force is the supportive force exerted by a surface perpendicularly to balance the weight of an object resting on it. Here, the ground supplies the normal force to the base of the pole.

• Normal force acts perpendicular to the contact surface and is essential for maintaining equilibrium.
• It counteracts the pole’s weight to keep it from moving downward through the ground.
• In this exercise, the normal force was calculated by balancing vertical forces, giving us a value of 2036 N.

The concept of normal force helps elucidate why objects don’t fall through the surfaces upon which they rest, and it's a vital part of the force balance in this exercise.