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Chapter 4: Problem 74

A mass \(M\) hangs from a uniform rope of length \(L\) and mass \(m\) Find an expression for the rope tension as a function of the distance \(y\) measured downward from the top of the rope.

Short Answer

Expert verified
The tension in the rope at a height \( y \) from the top is given by the function \( T(y) = g m ( 1 - \frac{y}{L} ) + Mg \), where \( g \) is the acceleration due to gravity, \( m \) and \( M \) are the masses of the rope and hanging load respectively, and \( L \) is the length of the rope.

Step by step solution

01

Identify Forces Acting on the Section

First, visualize the scenario and identify the forces. There are two forces acting on the section of rope located at a distance \( y \) from the top of the rope: the tension \( T(y) \) at the top and the weight of the rope's part and the mass \( M \) hanging below the point.
02

Represent the Rope Weight

Next, determine the weight of the part of the rope below the point. The weight density of the rope can be calculated as \( \lambda = \frac{m}{L} \), which represents the weight per unit length. The part of the rope below a point \( y \) has a length of \( L-y \) and thus a weight of \( g \lambda (L-y) = g \frac{m}{L} (L-y) = g m ( 1 - \frac{y}{L} ) \), where \( g \) is the acceleration due to gravity.
03

Use Newton's Second Law

Apply Newton's second law, which states that the sum of the forces acting on an object equals its mass times its acceleration. The forces acting on the section of the rope are its weight downward (\( g m ( 1 - \frac{y}{L} ) + Mg \)) and the rope tension upward (\( -T(y) \)). As the rope hangs vertically and at rest, the acceleration of the section is zero. This leads to the equation: \( T(y) = g m ( 1 - \frac{y}{L} ) + Mg \).

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