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Chapter 7: Problem 3

A \(120-\mathrm{kg}\) mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. \(\mathrm{A}\) postal worker then displaces the bag to a position 2.0 \(\mathrm{m}\) sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

Short Answer

Expert verified
(a) 233 N; (b)(i) 0 J, (ii) 466 J.

Step by step solution

01

Determine the Triangle Dimensions

The bag hangs initially straight down. When it's moved sideways, the rope forms a right triangle with the vertical and horizontal components. Use the Pythagorean theorem: the rope length is 3.5 m (hypotenuse), and it is described as moved 2.0 m horizontally. Find the vertical component: \[ h = \sqrt{3.5^2 - 2.0^2} = \sqrt{12.25 - 4.0} = \sqrt{8.25} \approx 2.87 \, \text{m}.\]
02

Calculate the Tension Force in the Rope

The tension in the rope, T, must support the weight of the bag vertically while being displaced horizontally. Use trigonometry to find the tension:\[T \cdot \cos(\theta) = mg \, \text{where} \, m = 120 \text{ kg}, \, g = 9.8 \, \text{m/s}^2.\]Given \( h = 2.87 \, \text{m} \) and hypotenuse = 3.5 m, find \( \cos(\theta) = 2.87/3.5 \), thus:\[T = \frac{mg}{\cos(\theta)} = \frac{120 \times 9.8}{2.87/3.5} \approx 408 \text{ N}.\]
03

Solve for the Horizontal Force

The horizontal force is the horizontal component of the tension needed to keep the bag stationary:\[F_{horizontal} = T \cdot \sin(\theta) = T \cdot \left( \frac{2.0}{3.5} \right).\]Using the tension found, \[F_{horizontal} = 408 \times \left(\frac{2.0}{3.5}\right) \approx 233 \, \text{N}.\]
04

Calculate Work Done by the Rope

Work (W) is calculated as force times displacement in the direction of the force. Since the force by the rope is perpendicular to the displacement of the motion (V = 2.0 m, side to side), \[W_{rope} = T \cdot d \cdot \cos(90^\circ) = 0.\]The work done by the rope is zero because the tension force does no work as it acts perpendicular to the bag's displacement.
05

Calculate Work Done by the Worker

The work done by the worker is equal to the horizontal force exerted over the distance:\[W_{worker} = F_{horizontal} \times d = 233 \, \text{N} \times 2.0 \, \text{m} = 466 \, \text{J}.\]

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Key Concepts

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

Trigonometry in Physics
When solving physics problems involving angles and distances, trigonometry becomes a crucial tool. In this problem, the bag hanging by a cord forms a right triangle when displaced sideways. The hypotenuse of this triangle is the length of the rope, which remains constant at 3.5 meters. This triangle setup allows us to use the Pythagorean theorem and trigonometric identities such as sine and cosine to find unknowns.
  • To find the vertical component (height) when the bag is displaced, we apply the Pythagorean theorem. We calculate: \[ h = \sqrt{3.5^2 - 2.0^2} = \sqrt{8.25} \approx 2.87 \text{ m}.\]

  • To relate this to the forces, we need to calculate the angles using sin and cos functions. Here, \[ \cos(\theta) = \frac{2.87}{3.5} \]

With these mathematical principles, we identify how forces act in different directions based on angles and known lengths.
Work and Energy
The concept of work and energy in physics explains how force causes displacement. Work is defined as the product of the force applied and the displacement in the direction of that force. This exercise demonstrates how energy is transferred via work done by forces like tension and a worker.
  • The tension in the rope that suspends the bag does no work. This is because the force from the rope is perpendicular to the displacement of the bag (90 degrees), resulting in zero work: \[ W_{rope} = T \cdot d \cdot \cos(90^\circ) = 0\]

  • Conversely, the worker applies a horizontal force to move the bag sideways, parallel to the displacement. The work done by this force is calculated using: \[ W_{worker} = F_{horizontal} \times d = 233 \text{ N} \times 2.0 \text{ m} = 466 \text{ J} \]

In sum, the exertion of a force in the direction of movement directly translates into the work done, measured in joules (J). This concept of work intertwines directly with the concept of energy, as work done translates into energy used or transferred.
Forces and Tension
Analyzing forces and tension is essential in understanding the dynamics of a system in equilibrium. In this scenario, the tension in the rope arises from the weight of the 120 kg bag, which is supported while being held at an angle. The vertical component of tension balances the weight, while the horizontal component balances the horizontal force applied.
  • To determine the overall tension in the rope, we consider the equilibrium of forces. The vertical component of the tension equals the gravitational force: \[ T \cdot \cos(\theta) = mg\] We solve for tension, T: \[ T = \frac{mg}{\cos(\theta)}\]

  • The horizontal component of tension equates to the force needed to hold the bag in its displaced position. Using the sine function, we solve: \[ F_{horizontal} = T \cdot \sin(\theta) = T \cdot \left( \frac{2.0}{3.5} \right) \]

This breakdown highlights how tension not only supports the weight vertically but helps establish the system's balance when altered by horizontal movements. Understanding these forces aids in analyzing similar equilibrium problems in physics.

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Most popular questions from this chapter

Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended 1.20 \(\mathrm{m}\) , its speed is 3.00 \(\mathrm{m} / \mathrm{s}\) . If the total mass of the two blocks is \(15.0 \mathrm{kg},\) what is the mass of each block?

A hydroelectric dam holds back a lake of surface area \(3.0 \times 10^{6} \mathrm{m}^{2}\) that has vertical sides below the water level. The water level in the lake is 150 \(\mathrm{m}\) above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted into electrical energy with 90\(\%\) efficiency. (a) If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is 1000 \(\mathrm{kg} / \mathrm{m}^{3} .(\mathrm{b})\) What volume of water must pass through the dam to produce 1000 kilo-watt-hours of electrical energy? What distance does the level of water in the lake fall when this much water passes through the dam?

A stone of mass \(m\) is thrown upward at an angle \(\theta\) above the horizontal and feels no appreciable air resistance. Use conservation of energy to show that at its highest point, it is a distance \(v_{0}^{2}\left(\sin ^{2} \theta\right) / 2 g\) above the point where it was launched. (Hint: \(v_{0}^{2}=v_{\mathrm{ax}}^{2}+v_{\mathrm{oy}}^{2} . )\)

A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible. (a) Calculate the potentinl-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?

Bungee Jump. A bungee cord is 30.0 \(\mathrm{m}\) long and, when stretched a distance \(x,\) it exerts a restoring force of magnitude \(k x\) Your father-in- law (mass 95.0 \(\mathrm{kg}\) ) stands on a platform 45.0 \(\mathrm{m}\) above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only 41.0 \(\mathrm{m}\) before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of 380.0 \(\mathrm{N}\) . When you do this, what distance will the bungee cord that you should select have stretched?
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