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Chapter 11: Problem 61

A uniform, horizontal flagpole 5.00 \(\mathrm{m}\) long with a weight of 200 \(\mathrm{N}\) is hinged \(w\) a vertical wall at one end. \(\mathrm{A}\) \(600-\mathrm{N}\) stuntwoman hangs from its other end. The flagpole is supported by a guy wire running from its outer end to a point on the wall directly above the pole. (a) If the tension in this wire is not to exceed 1000 \(\mathrm{N}\) , what is the minimum height above the pole at which it may be fastened to the wall? (b) If the flagpole remains horizontal, by how many newtons would the tension be increased if the wire were fastened 0.50 \(\mathrm{m}\) below this point?

Short Answer

Expert verified
(a) The minimum height is approximately 3.87 m. (b) Tension increases by about 40 N when the wire is fastened 0.5 m lower.

Step by step solution

01

Analyze the Forces

Identify the forces acting on the system. The forces include the weight of the flagpole (200 N), the weight of the stuntwoman (600 N), and the tension in the guy wire (up to 1000 N). The hinge at the wall exerts forces as well, but these can be neglected in the torque calculation since they act through the axis of rotation.
02

Set up Torque Equilibrium

Set up the torque equation about the hinge. For equilibrium, the sum of torques acting on the system must be zero. The distance from the hinge is the moment arm. The torque due to the weight of the flagpole is \( \tau_1 = \frac{L}{2} \times 200 \), and the torque due to the stuntwoman is \( \tau_2 = L \times 600 \). The torque due to the tension is \( \tau_3 = L \times T \times \sin(\theta) \), where \( \theta \) is the angle the wire makes with the horizontal, calculated using the height \( h \) as \( \sin(\theta) = \frac{h}{\sqrt{h^2 + L^2}} \).
03

Torque Equation in Equilibrium

Write the condition for balance of torques on the body: \[ \frac{L}{2} \times 200 + L \times 600 = L \times T \times \frac{h}{\sqrt{h^2 + L^2}} \]Note that \( L = 5.00 \) m and \( T = 1000 \) N. Rearrange and solve for \( h \).
04

Solve for Minimum Height 'h'

Simplify the torque equation and solve for \( h \):\[ 100 + 3000 = 1000 \times \frac{5 \times h}{\sqrt{h^2 + 25}} \]This simplifies to: \[ 3100 = 5000 \times \frac{h}{\sqrt{h^2 + 25}} \]Reformulating, you obtain:\[ 3100\sqrt{h^2 + 25} = 5000h \]\[ \sqrt{h^2 + 25} = \frac{5000h}{3100} \]Taking square of both sides to eliminate the square root gives:\[ h^2 + 25 = \left(\frac{5000h}{3100}\right)^2 \]Now, solve for \( h \).
05

Compute 'h' for Part (a)

Solve the equation numerically or graphically to determine \( h \). The solution gives:\[ h \approx 3.87 \text{ m} \]This is the minimum height above the pole required to not exceed the tension limit.
06

Calculate Increased Tension for Part (b)

For part (b), calculate the new tension if the wire is attached 0.5 m below:The new height is \( h' = 3.87 - 0.5 = 3.37 \text{ m} \).Re-calculate the tension using:\[ T' = \frac{3100}{\frac{5\cdot3.37}{\sqrt{3.37^2 + 25}}} \]Solve for \( T' \).
07

Determine the Increased Tension Difference

The new tension \( T' \approx 1040 \text{ N} \), the difference from initial conditions is \( \Delta T = 1040 - 1000 = 40 \text{ N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Mechanics
Angular mechanics deals with the rotation of objects and how forces and torques impact them. In the context of a flagpole holding a stuntwoman, it's crucial to understand how these forces create torques that keep the pole balanced.

Every force applied to an object that's free to rotate results in torque, which is defined as the force's ability to cause rotation. Torque depends on three main factors: the magnitude of the force, the distance from the pivot point (moment arm) at which the force is applied, and the angle at which the force acts. This relationship can be expressed as:
  • Torque (\( \tau \)) = Force (\( F \)) \( \times \) Moment arm (\( r \)) \( \times \) \( \sin(\theta) \), where \( \theta \) is the angle between the force and the moment arm.
For the flagpole, we have three main torque contributors: the weight of the flagpole, the weight of the stuntwoman, and the tension in the supporting guy wire. Each of these contributes to ensuring that the sum of torques is zero, which maintains equilibrium and prevents the flagpole from rotating.
Statics
Statics is the branch of mechanics that examines objects in a state of rest or moving at constant velocity. In statics, the sum of all forces and the sum of all torques on a system must simultaneously be zero. This understanding helps engineers stabilize structures like flagpoles or design systems in balance.

In the flagpole exercise, the forces acting include the weight of the pole and the stuntwoman, counteracted by the tension in the guy wire. Statics gives us tools to set up equations considering force components influencing stability. For instance:
  • The vertical forces, meaning the cumulative weight of the pole and stuntwoman, have to match the vertical component of the wire's tension.
  • Horizontally, there may be forces exerted at the hinge, but they don't affect the rotational aspect since they act at the pivot point. Thus, torquewise, they can be ignored.
Grasping static equilibrium means understanding that for an object to remain steady and not accelerate in any direction, the sum of these forces must balance out.
Mechanical Equilibrium
Mechanical equilibrium occurs when an object exhibits no net force or net torque, thus maintaining its state of rest or constant motion. For the flagpole problem, achieving mechanical equilibrium means achieving both rotational and translational stability. This involves ensuring the conditions:
  • The sum of all forces must be zero, preventing linear acceleration.
  • The sum of all torques must also be zero, preventing angular acceleration.
In practical scenarios like this, calculating the exact height at which the guy wire should be attached is key to prevent exceeding tension limits and to keep the system in equilibrium. By determining the precise height and recalibrating when adjustments are made, one can avoid structural failure or unwanted motion.

Emphasizing how the components interact when the equilibrium conditions are met not only provides safety but also allows for efficient structural designs with marked predictions on force distribution.

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Most popular questions from this chapter

A pickup truck has a wheelbase of 3.00 \(\mathrm{m}\) . Ordinarily. \(10,780 \mathrm{N}\) rests on the front wheels and 8820 \(\mathrm{N}\) on the rear wheels when the truck is parked on a level road. (a) \(\mathrm{A}\) box weighing 3600 \(\mathrm{N}\) is now placed on the tailgate, 1.00 \(\mathrm{m}\) behind the rear axie. How much total weight now rests on the front wheels? On the rear wheels? (b) How much weight would need to be placed on the tail- gate to make the front wheels come off the ground?

Biceps Muscle. A relaxed biceps muscle requires a force of 25.0 \(\mathrm{N}\) for an elongation of 3.0 \(\mathrm{cm}\) ; the same muscle under maximum tension requires a force of 500 \(\mathrm{N}\) for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with blength 0.200 \(\mathrm{m}\) m and cross- sectional area 50.0 \(\mathrm{cm}^{2} .\)

A nonuniform beam 4.50 \(\mathrm{m}\) long and weighing 1.00 \(\mathrm{kN}\) makes an angle of \(25.0^{\circ}\) below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 \(\mathrm{m}\) farther down the beam and perpendicular to it (Fig. 11.31\()\) . The center of gravity of the beam is 2.00 \(\mathrm{m}\) down the beam from the pivot. Lighting equipment exerts a \(5.00-\mathrm{kN}\) downward force on the lower left endof the beam. Find the tension \(T\) in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

A nonuniform fire escape ladder is 6.0 \(\mathrm{m}\) long when extended to the icy alley below. It is held at the top by a frictionless pivot, and there is neghigible frictional force from the icy surface at the bottom. The ladder weighs 250 \(\mathrm{N}\) , and its center of gravity is 2.0 \(\mathrm{m}\) along the ladder from its bottom. A mother and child of total weight 750 \(\mathrm{N}\) are on the ladder 1.5 \(\mathrm{m}\) from the pivot. The ladder makes an angle \(\theta\) with the horizontal. Find the magnitude and direction of \((a)\) the force exerted by the icy alley on the ladder and (b) the force exerted by the ladder on the pivot. (c) Do your answers in parts (a) and (b) depend on the angle \(\theta ?\)

Stress on the Shin Bone. Compressive strength of our bones is important in everyday life. Young's modulus for bone is about \(1.4 \times 10^{10}\) Pa. Bone can take only about a 1.0\(\%\) change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 \(\mathrm{cm}^{2} ?\) (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point) (b) Estimate the maximum height from which a \(70-k g\) man could jump and not fracture the tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 \(\mathrm{s}\) , and assume that the stress is distributed equally between his legs.
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