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Expert-verified Found in: Page 2 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Question: The system in Fig. 12-28 is in equilibrium, with the string in the center exactly horizontal. Block A weighs 40 N , block B weighs 50 N , and angle ${\mathbit{\varphi }}$ is 350 . Find (a) tension T1 , (b) tension T2 , (c) tension T3 , and (d) angle ${\mathbit{\theta }}$ . 1. The tension T1 in the string is 49 N
2. The tension T2 in the string is 28 N
3. The tension T3 in the string is 57 N
4. The angle $\theta$ made by T3 with vertical is 290 .
See the step by step solution

## Step 1: Understanding the given information

The weight of the block A, WA = 40 N

The weight of the block B, WB =50 N

The angle made by T1 with vertical is, $\varphi ={35}^{0}$

## Step 2: Concept and formula used in the given question

Using the free body diagram and the condition for equilibrium, you can find the tensions in the string and the angle theta. The formulas used are given below.

Newton’s second law:

Fnet = ma

At equilibrium,

Fnet = 0

## Step 3: (a) Calculation for the tension  T1

The free body diagram: At equilibrium,

Fynet = 0

From the figure we can write,

$\\mathrm{begingathered}{W}_{A}={T}_{1}\text{cos}\varphi \phantom{\rule{0ex}{0ex}}{T}_{1}=\frac{{W}_{A}}{\text{cos}\varphi }\phantom{\rule{0ex}{0ex}}=\frac{40}{\text{cos}35°}\phantom{\rule{0ex}{0ex}}=49\text{N}\phantom{\rule{0ex}{0ex}}\\mathrm{endgathered}$

Hence, the tension T1 in the string is 49 N

## Step 3: (b) Calculation for the tension  T2

At equilibrium,

${F}_{xnet}=0\phantom{\rule{0ex}{0ex}}{T}_{2}={\text{T}}_{1}\text{sin}\varphi \phantom{\rule{0ex}{0ex}}=49\text{sin}35°\phantom{\rule{0ex}{0ex}}=28\text{N}\phantom{\rule{0ex}{0ex}}$

Hence, the tension T2 in the string is 28 N

## Step 3: (c) Calculation for the tension T3

From the free body diagram, you can write that: ${T}_{3x}={T}_{3}\text{sin}\theta \phantom{\rule{0ex}{0ex}}={T}_{2}\phantom{\rule{0ex}{0ex}}=28\text{N}\phantom{\rule{0ex}{0ex}}$

Also,${T}_{3y}={T}_{3}\text{cos}\theta \phantom{\rule{0ex}{0ex}}={\text{W}}_{\text{B}}\phantom{\rule{0ex}{0ex}}=50\text{N}\phantom{\rule{0ex}{0ex}}$

So, the tension T3 is,

${T}_{3}=\sqrt{{{T}_{3x}}^{2}+{{T}_{3y}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{28}^{2}+{50}^{2}}\phantom{\rule{0ex}{0ex}}=57.3\phantom{\rule{0ex}{0ex}}~57\text{N}\phantom{\rule{0ex}{0ex}}$

Hence, the tension T3 in the string is 57 N

## Step 3: (d) Calculation for the angle  θ

${T}_{3}\text{sin}\theta ={T}_{2}\phantom{\rule{0ex}{0ex}}57\text{sin}\theta =28\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{sin}}^{-1}\left(\frac{28}{57}\right)\phantom{\rule{0ex}{0ex}}\theta =29\phantom{\rule{0ex}{0ex}}$

Hence, the angle($\theta$) made by T3 with vertical is 290 . ### Want to see more solutions like these? 