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Q25 PE

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Found in: Page 317

### College Physics (Urone)

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

# 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 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 $${\rm{897}}\;{\rm{N}}$$.