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Q25 PE

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College Physics (Urone)
Found in: Page 317
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

Repeat Exercise \({\rm{9}}{\rm{.24}}\) for the pulley shown in Figure \({\rm{9}}{\rm{.26}}\) (c), assuming you pull straight up on the rope. The pulley system’s mass is \({\rm{7}}{\rm{.00}}\;{\rm{kg}}\).

The tension in the rope is

  1. \({\rm{299}}\;{\rm{N}}\).
  2. The ceiling force is \({\rm{897}}\;{\rm{N}}\).
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Force

A force is an external factor that can change the rest or motion of a body. It has a size and a general direction.

Step 2: Diagram

The diagram is given below:

The mass of the pulley is \({m_p} = 7\;{\rm{kg}}\).

Step 3: Calculation of the tension

(a)

According to the first condition of equilibrium, the net force on the system is zero.

The mass of the pulley and the engine is,

\(\begin{align}m &= 115 + 7\\ &= 122\;kg\end{align}\)

So, we can write,

\(\begin{align}mg &= 4T\\T &= \frac{{mg}}{4}\end{align}\)

For,\(m = 122\;{\rm{kg}}\) we write,

\(\begin{align}T &= \frac{{122 \times 9.8}}{4}\\ &= 299\;N\end{align}\)

Hence, the tension is \({\rm{299}}\;{\rm{N}}\).

Step 4: Calculation of the ceiling force

(b)

Assuming the rope is straight down and the ceiling force be \({{\rm{F}}_{\rm{c}}}\) we write,

\(\begin{align}{F_c} &= 3T\\ &= 3 \times 299\\ &= 897\;N\end{align}\)

Hence, the force is \({\rm{897}}\;{\rm{N}}\).

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