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Chapter 5: Problem 41

Using a rope that will snap if the tension in it exceeds \(387 \mathrm{~N}\), you need to lower a bundle of old roofing material weighing \(449 \mathrm{~N}\) from a point \(6.1 \mathrm{~m}\) above the ground. Obviously if you hang the bundle on the rope, it will snap. So, you allow the bundle to accelerate downward. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?

Short Answer

Expert verified
(a) The acceleration is approximately \(1.35 \mathrm{~m/s^2}\). (b) The speed on impact is approximately \(4.06 \mathrm{~m/s}\).

Step by step solution

01

Understanding the Forces

The forces acting on the bundle are gravity (its weight) and the tension in the rope. Since the weight of the bundle is greater than the tension, it will accelerate downward.
02

Setup the Equation for Forces

Using Newton's second law, we set up the equation: \[ F_{ ext{net}} = m imes a = W - T \]where \( W = 449 \, \mathrm{N} \) is the weight of the bundle, \( T = 387 \, \mathrm{N} \) is the tension, and \( a \) is the acceleration. Since both forces are in newtons, they can be directly compared.
03

Rearrange the Equation for Acceleration

Substitute the known values into the equation to solve for acceleration: \[ 449 \, N - 387 \, N = m imes a \].
04

Calculate Mass from Weight

Calculate the mass \( m \) of the bundle using its weight \( W = m imes g \) where \( g = 9.81 \, \mathrm{m/s^2} \). \[ m = \frac{W}{g} = \frac{449}{9.81} \approx 45.76 \, \mathrm{kg} \].
05

Solve for Acceleration

Using the mass from Step 4 in the net force equation from Step 3:\[ 449 - 387 = 45.76 imes a \]\[ 62 = 45.76 imes a \]\[ a \approx \frac{62}{45.76} \approx 1.35 \, \mathrm{m/s^2} \].
06

Calculate Final Speed with Acceleration

We use the kinematic equation \( v^2 = u^2 + 2a s \) where initial speed \( u = 0 \), acceleration \( a = 1.35 \, \mathrm{m/s^2} \), and distance \( s = 6.1 \, \mathrm{m} \). \[ v^2 = 0 + 2 \times 1.35 \times 6.1 \]\[ v^2 = 16.47 \]\[ v \approx \sqrt{16.47} \approx 4.06 \, \mathrm{m/s} \].
07

Conclusion

With an acceleration of approximately \(1.35 \mathrm{~m/s^2}\), the rope is on the verge of snapping, and the bundle will hit the ground at a speed of approximately \(4.06 \mathrm{~m/s}\).

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Key Concepts

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

Tension in the Rope
Tension in a rope refers to the pulling force transmitted along the rope, wire, or cable when it is used to transmit force. In this exercise, we are tasked to measure the tension that the rope can endure without snapping while lowering a bundle of old roofing material.

The bundle has a weight of 449 N, but the rope can only hold up to 387 N before breaking. By allowing the bundle to accelerate downward, we can manipulate the forces so that the tension in the rope reaches its threshold but does not surpass it.

Using Newton's second law, the balance between the bundle's weight and the tension determines the net force. This is pivotal as it allows us to calculate the exact acceleration that maximizes tension without breaking the rope, keeping us on the verge of snapping but not exceeding it.
Kinematics Equations
Kinematics equations help us understand motion details, such as the final velocity of a moving object, given specific conditions. They are essential in predicting how an object will behave when subjected to different forces and motions.

For the roofing material example, we used the equation \[v^2 = u^2 + 2a s\]where:
  • \(v\) is the final velocity.
  • \(u\) is the initial velocity (which is 0 since the bundle starts from rest).
  • \(a\) is the acceleration, calculated to be 1.35 \(\mathrm{m/s^2}\).
  • \(s\) is the distance fallen, 6.1 meters.
Plugging in these values, the equation helps us calculate the final speed of the bundle when it hits the ground, ensuring we understand how the forces and motion play out practically.
Weight and Mass Calculations
Calculating weight and mass is crucial in physics for understanding how different objects experience forces. In this exercise, the weight of the bundle is given as 449 N.

Weight is the force exerted by gravity on an object, and it is calculated using the formula:\[W = m \times g\]where:
  • \(W\) is the weight, 449 N.
  • \(m\) is the mass.
  • \(g\) is the acceleration due to gravity, approximately 9.81 \(\mathrm{m/s^2}\).
To find the mass of the bundle, we rearrange the equation:\[m = \frac{W}{g}\]Substitute the given weight and solve for mass to get approximately 45.76 kg.

Understanding the mass and weight relationship helps in computing other critical dynamics like acceleration, crucial for maintaining the rope's integrity while lowering heavy materials.

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Most popular questions from this chapter

A certain particle has a weight of \(22 \mathrm{~N}\) at a point where \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\). What are its (a) weight and (b) mass at a point where \(g=4.9 \mathrm{~m} / \mathrm{s}^{2} ?\) What are its (c) weight and (d) mass if it is moved to a point in space where \(g=0\) ?

A worker drags a crate across a factory floor by pulling on a rope tied to the crate. The worker exerts a force of magnitude \(F=450 \mathrm{~N}\) on the rope, which is inclined at an upward angle \(\theta=38^{\circ}\) to the horizontal, and the floor exerts a horizontal force of magnitude \(f=125 \mathrm{~N}\) that opposes the motion. Calculate the magnitude of the acceleration of the crate if (a) its mass is \(310 \mathrm{~kg}\) and \((\mathrm{b})\) its weight is \(310 \mathrm{~N}\).

A \(0.150 \mathrm{~kg}\) particle moves along an \(x\) axis according to \(x(t)=-13.00+2.00 t+4.00 t^{2}-3.00 t^{3}\), with \(x\) in meters and \(t\) in seconds. In unit-vector notation, what is the net force acting on the particle at \(t=3.40 \mathrm{~s}\) ?

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m}\). He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)

Holding on to a towrope moving parallel to a frictionless ski slope, a \(50 \mathrm{~kg}\) skier is pulled up the slope, which is at an angle of \(8.0^{\circ}\) with the horizontal. What is the magnitude \(F_{\text {rope }}\) of the force on the skier from the rope when (a) the magnitude \(v\) of the skier's velocity is constant at \(2.0 \mathrm{~m} / \mathrm{s}\) and \((\mathrm{b}) v=2.0 \mathrm{~m} / \mathrm{s}\) as \(v\) increases at a rate of \(0.10 \mathrm{~m} / \mathrm{s}^{2} ?\)
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