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Chapter 9: Problem 4

Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angle of \(27.0^{\circ}\) above the horizontal. One child, with a mass of \(44.0 \mathrm{kg},\) is hanging \(1.30 \mathrm{m}\) along the branch from the tree trunk. The other child, with a mass of \(35.0 \mathrm{kg}\), is hanging \(2.10 \mathrm{m}\) from the tree trunk. What is the magnitude of the net torque exerted on the branch by the children? Assume that the axis is located where the branch joins the tree trunk and is perpendicular to the plane formed by the branch and the trunk.

Short Answer

Expert verified
The net torque is approximately 579.9 Nm.

Step by step solution

01

Understand the Problem

We need to calculate the net torque exerted by two children hanging from a tree branch. The torque depends on the force applied (due to weight) and the distance from the pivot point (where the branch meets the trunk). The force due to each child's weight and the distances they hang from the trunk will be used to find individual torques, and then summed to find the net torque.
02

Calculate Force due to Weight

The force due to the weight of each child is calculated using \[ F = mg \]where \( m \) is mass, and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. - For the first child (44 kg): \[ F_1 = 44 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 431.2 \, \text{N} \]- For the second child (35 kg): \[ F_2 = 35 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 343.0 \, \text{N} \]
03

Determine Torque

Torque is calculated using the formula:\[ \tau = rF \sin(\theta) \]where \( r \) is the distance from the pivot, \( F \) is the force, and \( \theta \) is the angle between the branch and horizontal.- For the first child: \[ \tau_1 = 1.30 \, \text{m} \times 431.2 \, \text{N} \times \sin(27^{\circ}) \approx 254.0 \, \text{Nm} \]- For the second child: \[ \tau_2 = 2.10 \, \text{m} \times 343.0 \, \text{N} \times \sin(27^{\circ}) \approx 325.9 \, \text{Nm} \]
04

Calculate Net Torque

The net torque is the sum of the individual torques:\[ \text{Net Torque} = \tau_1 + \tau_2 \]\[ \text{Net Torque} = 254.0 \, \text{Nm} + 325.9 \, \text{Nm} = 579.9 \, \text{Nm} \]
05

Present the Final Answer

The magnitude of the net torque exerted on the branch by the children is approximately \( 579.9 \, \text{Nm} \).

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Key Concepts

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

Torque Formula
Torque is a measure of the rotational force that is applied about an axis. It is through this concept that we understand how forces cause objects to rotate. Ostensibly, torque is an essential element in understanding branch mechanics, especially when calculating the rotational effect caused by weights hanging from a branch.
The basic formula for torque is given by:
  • \[ \tau = rF \sin(\theta) \] This formula states that torque \( \tau \) is the product of three elements:
    • \( r \): the distance from the axis of rotation (pivot point) to the point where the force is applied.
    • \( F \): the magnitude of the force applied.
    • \( \theta \): the angle between the line of action of the force and the line from the pivot to the point of application.
Thus, torque is directly proportional to the distance from the pivot and the force applied, as well as the sine of the angle involved. In branch mechanics, it's important to consider this entire equation to deduce how much rotational effect a certain force, like a child's weight, will apply on a branch.
Force Due to Gravity
The force due to gravity is a crucial concept in physics, commonly referred to as weight. This force acts downwards due to the Earth's gravitational field.
In our branch scenario, the force due to gravity acting on each child is calculated using the formula:
  • \[ F = mg \] Here, \( m \) represents the mass of the object, and \( g \) is the gravitational acceleration, approximately \( 9.8 \, \text{m/s}^2 \).
For each child hanging from a branch, their weight (force due to gravity) pulls downwards on the branch.
  • The 44 kg child's force due to gravity is \( 431.2 \, \text{N} \).
  • The 35 kg child's force is \( 343.0 \, \text{N} \).
These forces play a pivotal role in calculating the torque exerted on the branch. Each child's weight calculates into a part of the total force applied in the torque formula.
Branch Mechanics
Branch mechanics involve analyzing the forces and torques acting on a branch, which helps in understanding real-life problems like the one where children hang from a branch.
In this scenario, we apply the principles of mechanics to determine how a tree branch handles the load of the children. Consider the following:
  • The branch grows out at an angle of \( 27.0^{\circ} \) with respect to the horizontal. This angle is important because it affects the torque calculation.
  • By knowing where the children hang, which is along this angle, we understand how their weights create a rotational effect about the axis where the branch meets the trunk.
  • The combined torques from the children's weights determine the branch's behavior 鈥� any excess may cause breaking or bending 鈥� a critical aspect in the physics of structures.
By assessing these mechanics, we derive not only the net torque but also appreciate how forces and structures interact, providing a deeper understanding of stability and equilibrium in physical systems.

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Most popular questions from this chapter

A clay vase on a potter's wheel experiences an angular acceleration of \(8.00 \mathrm{rad} / \mathrm{s}^{2}\) due to the application of a \(10.0-\mathrm{N} \cdot \mathrm{m}\) net torque. Find the total moment of inertia of the vase and potter's wheel.

A helicopter has two blades (see Figure 8.11 ); each blade has a mass of \(240 \mathrm{kg}\) and can be approximated as a thin rod of length \(6.7 \mathrm{m} .\) The blades are rotating at an angular speed of \(44 \mathrm{rad} / \mathrm{s}\). (a) What is the total moment of inertia of the two blades about the axis of rotation? (b) Determine the rotational kinetic energy of the spinning blades.

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about \(1.1 \times 10^{20} \mathrm{J}\)

Conceptual Example 13 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of \(125 \mathrm{kg} \cdot \mathrm{m}^{2} .\) When one person is standing on the carousel at a distance of \(1.50 \mathrm{m}\) from the center, the carousel has an angular velocity of \(0.600 \mathrm{rad} / \mathrm{s} .\) However, as this person moves inward to a point located \(0.750 \mathrm{m}\) from the center, the angular velocity increases to \(0.800 \mathrm{rad} / \mathrm{s} .\) What is the person's mass?

The wheel of a car has a radius of \(0.350 \mathrm{m}\). The engine of the car applies a torque of \(295 \mathrm{N} \cdot \mathrm{m}\) to this wheel, which does not slip against the road surface. since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?
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