/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 One end of a \(30-\mathrm{m}\) r... [FREE SOLUTION] | 91Ó°ÊÓ

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One end of a \(30-\mathrm{m}\) rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls sideways on the midpoint of the rope, displacing it a distance of \(2 \mathrm{m}\). If he exerts a force of \(80 \mathrm{N}\) under these conditions, determine the force exerted on the car.

Short Answer

Expert verified
The force exerted on the car by the rope is approximately \(156.88 N\).

Step by step solution

01

Understand the setup and define the variables

Here, we can define the following variables: - Total length of the rope (L) = 30 m. - Displacement of the midpoint of the rope (D) = 2 m. - Force exerted by the motorist (F) = 80 N. - The force exerted on the car (F_car) is what we want to determine.
02

Identify the triangle and split it into two right-angled triangles

We can visualize the problem by drawing the triangle formed by the tree, the car, and the displaced midpoint of the rope. Since the motorist pulls at the midpoint of the rope, we can split the triangle into two right-angled triangles where the hypotenuse of both triangles is equal to half the length of the rope (L/2 = 15 m).
03

Calculate the angles of the right-angled triangles

Now, let's denote the angle between the hypotenuse and the base side (adjacent to the angle) of the triangle as \(\theta\). Using trigonometric ratios, we can find the angle \(\theta\): \(\sin(\theta) = \frac{opposite}{hypotenuse}\) Since the opposite side is the distance (D) and the hypotenuse is half the total length of the rope (L/2), we can plug in the values: \(\sin(\theta) = \frac{2}{15}\) Now, we can find the angle: \(\theta = \arcsin(\frac{2}{15}) \approx 7.72^\circ\)
04

Calculate the horizontal component of the force exerted by the rope

To calculate the force exerted on the car, we need to determine the horizontal component of the force exerted by the rope (F_horizontal). Using trigonometry, we can find the horizontal component by multiplying the force exerted by the motorist (F) with the cosine of the angle theta: \(F_{horizontal} = F \cdot \cos(\theta)\) Plugging in the values, we get: \(F_{horizontal} = 80 \cdot \cos(7.72^\circ) \approx 78.44 N\)
05

Calculate the force exerted on the car

Since there are two equal right-angled triangles, the force exerted by the rope will be split between them. Therefore, the total force exerted on the car can be calculated as the sum of the horizontal components of the force in these two triangles: \(F_{car} = 2 \cdot F_{horizontal}\) Substituting the value of F_horizontal: \(F_{car} = 2 \cdot 78.44 N \approx 156.88 N\) The force exerted on the car by the rope is approximately 156.88 N.

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Key Concepts

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

Trigonometric Ratios
The concept of trigonometric ratios is fundamental to understanding problems involving right-angled triangles. In its essence, these ratios provide a relationship between the angles and sides of a right-angled triangle. The three primary trigonometric ratios are sine (sin), cosine (cos), and tangent (tan), each relating to different pairs of sides of the triangle.

When faced with a physics problem that involves displacement and forces acting at an angle, such as pulling a car out of the mud with a rope, these ratios become invaluable. If we represent the angle of interest by \(\theta\), then these ratios are defined as follows:
  • \(\sin(\theta) = \frac{opposite}{hypotenuse}\)
  • \(\cos(\theta) = \frac{adjacent}{hypotenuse}\)
  • \(\tan(\theta) = \frac{opposite}{adjacent}\)
Using trigonometric ratios helps break down forces into components that are easier to analyze scientifically, thus allowing for a clearer and more precise force calculation.
Right-Angled Triangles
Right-angled triangles are the cornerstones of trigonometry. Every right-angled triangle, which includes a right angle (90 degrees), has two additional angles which always sum to 90 degrees. This characteristic is central to many physical situations where forces are not aligned or are acting at an angle.

The significance of right-angled triangles in the exercise given is exemplified by the fact that pulling the midpoint of the rope creates two congruent right-angled triangles with the ground serving as one side, the rope as the hypotenuse, and the displacement as the height. By utilizing properties of these triangles, we can calculate unknown forces and angles. It's worth noting that for right-angled triangles, Pythagoras' theorem also holds true, which states that the hypotenuse squared is equal to the sum of the squares of the other two sides: \(a^2 + b^2 = c^2\). Recognizing such triangles within a physics problem is a key step to solving it accurately.
Force Calculation
Force calculation in physics often requires understanding the direction and magnitude of forces. When using trigonometric ratios to calculate force, it becomes apparent that forces may not always act in straight lines or parallel to coordinate axes. As in the exercise, the force a motorist exerts on a rope is at an angle, meaning the force's effect will have components in both horizontal and vertical directions.

Once the angle \(\theta\) has been determined using trigonometry, we can proceed to find the horizontal component of the force exerted, which affects the car. By multiplying the magnitude of the exerted force (F) by the cosine of the angle (\(\cos(\theta)\)), we obtain the horizontal force component. In many physics problems, including the given exercise, it is essential to calculate the resultant force from various components. This often involves summing up vector quantities and employing vector addition techniques. Proper force calculation is imperative for, but not limited to, predicting the movement of objects, analyzing mechanical systems, and solving static equilibrium problems.

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Most popular questions from this chapter

When the Moon is directly overhead at sunset, the force by Earth on the Moon, \(F_{\mathrm{EM}},\) is essentially at \(90^{\circ}\) to the force by the Sun on the Moon, \(F_{\mathrm{SM}},\) as shown below. Given that \(F_{\mathrm{EM}}=1.98 \times 10^{20} \mathrm{N} \quad\) and \(F_{\mathrm{SM}}=4.36 \times 10^{20} \mathrm{N}, \quad\) all other forces on the Moon are negligible, and the mass of the Moon is \(7.35 \times 10^{22} \mathrm{kg}\) determine the magnitude of the Moon's acceleration.

A truck collides with a car, and during the collision, the net force on each vehicle is essentially the force exerted by the other. Suppose the mass of the car is \(550 \mathrm{kg}\), the mass of the truck is \(2200 \mathrm{kg}\), and the magnitude of the truck's acceleration is \(10 \mathrm{m} / \mathrm{s}^{2}\). Find the magnitude of the car's acceleration.

At a circus, a donkey pulls on a sled carrying a small clown with a force given by \(2.48 \hat{\mathbf{i}}+4.33 \hat{\mathbf{j}} \mathrm{N}\). A horse pulls on the same sled, aiding the hapless donkey, with a force of \(6.56 \hat{\mathbf{i}}+5.33 \hat{\mathbf{j}} \mathrm{N} .\) The mass of the sled is 575 kg. Using í and \(\hat{\mathbf{j}}\) form for the answer to each problem, find (a) the net force on the sled when the two animals act together, (b) the acceleration of the sled, and (c) the velocity after \(6.50 \mathrm{s}.\)

A ball of mass \(m\) hangs at rest, suspended by a string. (a) Sketch all forces. (b) Draw the free-body diagram for the ball.

Astronauts in orbit are apparently weightless. This means that a clever method of measuring the mass of astronauts is needed to monitor their mass gains or losses, and adjust their diet. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of \(50.0 \mathrm{N}\) is exerted, and an astronaut's acceleration is measured to be \(0.893 \mathrm{m} / \mathrm{s}^{2} .\) (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which she orbits experiences an equal and opposite force. Use this knowledge to find an equation for the acceleration of the system (astronaut and spaceship) that would be measured by a nearby observer. (c) Discuss how this would affect the measurement of the astronaut's acceleration. Propose a method by which recoil of the vehicle is avoided.

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