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Chapter 8: Problem 40

A uniform chain with a mass of \(1.32 \mathrm{~kg}\) per meter of length is coiled on a table. One end is pulled upward at a constant rate of \(0.47 \mathrm{~m} / \mathrm{s}\). a) Calculate the net force acting on the chain. b) At the instant when \(0.15 \mathrm{~m}\) of the chain has been lifted off the table, how much force must be applied to the end being raised?

Short Answer

Expert verified
Answer: The force applied to lift the chain at the specific instant is approximately 1.94 N.

Step by step solution

01

Write down the given information

We are given the following values: - mass per meter of the chain: \(1.32\mathrm{~kg/m}\) - speed with which the chain is being pulled upward: \(0.47\mathrm{~m/s}\) - length of chain lifted off the table at a specific instant: \(0.15\mathrm{~m}\)
02

Calculate the net force acting on the chain

To find the net force acting on the chain, we'll use the equation: \(F_{net} = m_{lifted}\cdot a\) Here, \(m_{lifted}\) is the mass of the chain that has been lifted off the table, and \(a\) is the acceleration due to gravity (which is \(9.81\mathrm{~m/s^2}\)). To find \(m_{lifted}\), we multiply the mass per meter of the chain by the length of the chain lifted off the table: \(m_{lifted} = (1.32\mathrm{~kg/m}) \cdot (0.15\mathrm{~m}) = 0.198\mathrm{~kg}\) Now, we can calculate the net force: \(F_{net} = 0.198\mathrm{~kg} \cdot 9.81\mathrm{~m/s^2} = 1.94\mathrm{~N}\) So the net force acting on the chain is approximately \(1.94\mathrm{~N}\).
03

Calculate the force applied to lift the chain at the specific instant

Since the chain is being pulled upward at a constant rate, the force required to lift the chain is equal to the net force acting on the chain. Thus, at the instant when \(0.15\mathrm{~m}\) of the chain has been lifted off the table, the force applied to lift the chain is also approximately \(1.94\mathrm{~N}\).

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