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Chapter 7: Problem 3

A 90.0 kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 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) The necessary horizontal force to hold the bag in the new position is 504 N. (b) The work done by the rope is 0 Joules, and the work done by the worker is 1008 Joules.

Step by step solution

01

Calculate the Resultant Force

Begin by calculating the hypotenuse of the triangle formed by the displacement of the bag using Pythagoras' theorem. The hypotenuse \( h \) equals to the square root of \( d^2 + L^2 \) where \( d \) is the displacement and \( L \) is the length of the rope. So \( h = \sqrt{(2.0 m)^2 + (3.5 m)^2} = 4.0 m \). The resultant tension force in the rope \( T \) is the weight divided by cosine of the angle between the rope and the vertical axis. The cosine of the angle \( cos(\theta) \) is \( L/h \), so \( T = (90.0 kg * 9.8 m/s^2) / cos(\theta) = 90.0kg * 9.8 m/s^2 * (4.0 m/3.5 m) = 1008 N \).
02

Find the Horizontal Force

The horizontal force \( F \) equals to \( T * sin(\theta) \). In triangle formed by the displacement of the bag, \( sin(\theta) = d/h \). So, \( F = T * sin(\theta) = 1008 N * (2.0 m/4.0 m) = 504 N \). This is the force that the postal worker must exert to hold the bag in place.
03

Calculate the Work Done by the Rope

The rope pulls the bag perpendicular to the displacement. Therefore, the angle \( \theta \) between the force exerted by the rope and the direction of displacement is 90 degrees. The formula of work done \( W = F*d*cos(\theta) \), thus work done by the rope \( W_{rope} = T*d*cos(90) = 0 Joules \) because \( cos(90) = 0 \).
04

Calculate the Work Done by the Worker

The work done by the worker is the product of the force exerted by the worker and the displacement of the bag along the direction of the force. So the work done by the worker \( W_{worker} = F*d = 504 N * 2.0 m = 1008 Joules \).

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Key Concepts

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

Newton's Laws
Newton's Laws of Motion are fundamental concepts in physics that describe how objects move under the influence of different forces. Let's break these down:

  • First Law (Law of Inertia): An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
  • Second Law: The acceleration of an object depends directly on the net force acting upon the object and inversely on the object's mass. This is described by the equation: \( F = ma \), where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration.
  • Third Law: For every action, there is an equal and opposite reaction.
In the context of the original problem, when the postal worker displaces the mail bag horizontally, they apply a force to counteract the horizontal tension created in the rope. The force necessary to hold the bag in place stems from the Second Law, as it matches the force exerted by tension horizontally, demonstrating Newton's principle of force equilibrium.
Work and Energy
Work and energy are closely related topics in physics. The concept of work in physics is defined as the product of the force applied to an object and the distance over which that force is applied. Work is done when a force moves an object.

The formula for calculating work is:
  • \( W = F \cdot d \cdot \cos(\theta) \)
Where:
  • \( W \) is work
  • \( F \) is the magnitude of the force
  • \( d \) is the displacement
  • \( \theta \) is the angle between the force and displacement direction
In the exercise, the work done by the rope is zero because the displacement is perpendicular to the direction of the tension force in the rope \((\cos(90) = 0)\). The worker, however, does work on the bag by applying a horizontal force over the horizontal displacement, calculated as 1008 Joules. This energy represents the orderly transfer and conversion of energy from the worker to the motion of the bag.
Trigonometry in Physics
Trigonometry is essential in physics, especially when dealing with forces acting at angles. Understanding how to resolve these forces into components can simplify the analysis of physical problems.

In the context of the mail bag problem:
  • The angle \( \theta \) between the rope and the vertical axis is critical for determining both the tension in the rope and the necessary horizontal force.
  • Tension \( T \) involves calculating the hypotenuse of the triangle formed by the rope's tilt, using the Pythagorean Theorem.
  • Using \( \sin(\theta) \) and \( \cos(\theta) \) allows us to project the forces onto the horizontal and vertical axes, facilitating the calculation of separate components.
Trigonometry helps in resolving complex forces and understanding their actions. By employing %sin(\theta)% and %cos(\theta)% we can meticulously break down the forces acting on any object in motion or at rest.

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Most popular questions from this chapter

CALC The potential energy of a pair of hydrogen atoms separated by a large distance \(x\) is given by \(U(x)=-C_{6} / x^{6},\) where \(C_{6}\) is a positive constant. What is the force that one atom exerts on the other? Is this force attractive or repulsive?

You are designing a delivery ramp for crates containing exercise equipment. The \(1470 \mathrm{~N}\) crates will move at \(1.8 \mathrm{~m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(515 \mathrm{~N}\) kinctic friction force on cach crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of \(5.0 \mathrm{~m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the largest force constant of the spring that will be needed to meet the design criteria.

A \(2.00 \mathrm{~kg}\) block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{~N} / \mathrm{m}\), compressing it \(0.220 \mathrm{~m}\). When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope \(37.0^{\circ}\) (Fig. \(\mathbf{P 7 . 4 0}\) ). (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

The food calorie, equal to \(4186 \mathrm{~J},\) is a measure of how much energy is released when the body metabolizes food. A certain fruit-and-cereal bar contains 140 food calories. (a) If a 65 kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical, only \(20 \%\) of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that \(100 \%\) of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.

A small rock with mass \(m\) is released from rest at the inside rim of a large, hemispherical bowl (point \(A\) ) that has radius \(R\), as shown in Fig. E7.9. If the normal force exerted on the rock as it slides through its lowest point (point \(B\) ) is twice the weight of the rock, how much work did friction do on the rock as it moved from \(A\) to \(B ?\) Express your answer in terms of \(m, R,\) and \(g\)
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