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Chapter 5: Problem 17

The distance between two telephone poles is \(50.0 \mathrm{m}\) When a \(1.00-\mathrm{kg}\) bird lands on the telephone wire midway between the poles, the wire sags \(0.200 \mathrm{m}\). Draw a free-body diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire.

Short Answer

Expert verified
The tension that the bird produces in the wire is equal to the weight of the bird divided by the sine of the angle between the tension and the horizontal.

Step by step solution

01

- Find the weight of the bird

First, find the weight of the bird using the formula \(F = m \cdot g\), where \(F\) is the force (weight), \(m\) is the mass of the bird which is \(1.00 kg\), and \(g\) is the acceleration due to gravity, which is around \(9.8 m/s^2\) on Earth. Substituting these values in, we find \(F = 1.00 \cdot 9.8 = 9.8N\). So, the force due to the bird's weight is \(9.8 N\).
02

- Establish the geometry of the problem

The sagging of the wire essentially forms a triangle with the bird at the vertex and the two telephone poles as the endpoints of the base of the triangle. We can see that this triangle is an isosceles triangle (equal sides) because the bird is exactly midway between the two poles. Now consider a half of this triangle. This is a right triangle given the wire from a pole to the bird forms the hypotenuse. Fore a half of this triangle, the base is \(25 m\) and the height is \(0.20 m\). Find the angle \(\alpha\) between the tension \(T\) and the horizontal by using the formula for a tangent in a right triangle \(tan(\alpha) = \frac{0.20m}{25m} --> \alpha = arctan(\frac{0.20m}{25m})\).
03

- Calculate tension in the wire

The total tension can be supposed to act along the hypotenuse of the triangle. Then tension can be found by using the equilibrium conditions in vertical and horizontal directions. The vertical component of tension cancels the bird's weight and the horizontal component is equal in both halves of the wire, essentially cancelling out. Thus \(\sigma_y: T\cdot sin(\alpha) = W\) and \(\sigma_x: 2 \cdot T\cdot cos(\alpha) = 0\). From \(\sigma_x\), we can see that \(T\cdot cos(\alpha)\) should equal zero. So the answer for the tension \(T\) can be found by \(T = W / sin(\alpha)\), where \(W\) is bird's weight force and \(\alpha\) is the angle we calculated in previous step. Replace \(W\) with \(9.8 N\) and \(\alpha\) with its calculated value and find the answer.

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Key Concepts

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

Free-body Diagram
Understanding the forces acting on an object is crucial, and that's where a free-body diagram comes into play. It's a simple representation of the object of interest, depicted as a point, and the forces acting on it, each vector coming off the point showing the direction and magnitude of the force.

For the problem at hand involving the bird on the telephone wire, we draw the free-body diagram by representing the bird as a point. From this point, we draw two forces: the weight of the bird acting downward due to gravity and the tension forces at an angle 鈥� because the wire sags, the tension isn't purely horizontal or vertical but diagonally upwards.
Equilibrium Conditions
The equilibrium conditions come into effect when an object is at rest or moving with constant velocity. This state can be described by two main conditions: no net force in the horizontal direction (the sum of all horizontal forces equals zero) and no net force in the vertical direction (the sum of all vertical forces equals zero).

In our exercise, the bird is stationary on the wire, implying that all forces on it must be balanced. The vertical forces are the weight of the bird and the upward components of tension from the wire, which must be equal in magnitude to the weight for equilibrium. The horizontal components of the tension cancel each other out because they are equal in magnitude and opposite in direction.
Triangle Geometry
Triangle geometry offers a way to visualize and solve many physics problems, particularly ones involving forces at angles. In this case, the sagging wire creates an isosceles triangle, which we halve to form two right-angled triangles.

By examining a half of this triangle, we gain an understanding of the relationship between the angle of the wire and the resultant tension. The tangent of this angle is directly related to the height (the sag of the wire) and the half-length of the wire between the pole and the bird. Using trigonometric identities and properties can give us a clearer picture of how the tension forces oppose the bird's weight.
Weight Force Calculation
The weight force calculation is a fundamental step in physics problems that involve gravity. It is determined by multiplying an object's mass by the acceleration due to gravity (usually represented by the symbol g and approximately equal to 9.8 m/s2 on Earth).

In our scenario, we calculate the bird's weight to be 9.8 N (newtons), which is a force vector pointing straight down. Recognizing that this weight force acts as the only vertical force helps us solve for tension using the geometry of the problem and the equilibrium conditions.

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Most popular questions from this chapter

You and your friend go sledding. Out of curiosity, you measure the constant angle \(\theta\) that the snow-covered slope makes with the horizontal. Next, you use the following method to determine the coefficient of friction \(\mu_{k}\) between the snow and the sled. You give the sled a quick push up so that it will slide up the slope away from you. You wait for it to slide back down, timing the motion. It turns out that the sled takes twice as long to slide down as it does to reach the top point in the round trip. In terms of \(\theta\), what is the coefficient of friction?

A Chevrolet Corvette convertible can brake to a stop from E speed of \(60.0 \mathrm{mi} / \mathrm{h}\) in a distance of \(123 \mathrm{ft}\) on a level roadway. What is its stopping distance on a roadway sloping downward at an angle of \(10.0^{\circ} ?\)

Draw a free-body diagram of a block which slides down a frictionless plane having an inclination of \(\theta=15.0^{\circ}\) (Fig. P5.22). The block starts from rest at the top and the length of the incline is \(2.00 \mathrm{m} .\) Find (a) the acceleration of the block and (b) its speed when it reaches the bottom of the incline. the block and (b) its speed when it reaches the bottom of the incline.

A 1.30 -kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal countertop is \(0.350 .\) To make the toaster start moving, you carelessly pull on its electric cord. (a) For the cord tension to be as small as possible, you should pull at what angle above the horizontal? (b) With this angle, how large must the tension be?

A force \(\mathbf{F}\) applied to an object of mass \(m_{1}\) produces an acceleration of \(3.00 \mathrm{m} / \mathrm{s}^{2} .\) The same force applied to a second object of mass \(m_{2}\) produces an acceleration of \(1.00 \mathrm{m} / \mathrm{s}^{2} .\) (a) What is the value of the ratio \(m_{1} / m_{2} ?\) (b) If \(m_{1}\) and \(m_{2}\) are combined, find their acceleration under the action of the force \(\mathbf{F}\).
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