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Q10P

Expert-verifiedFound in: Page 2

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**

**Answer:**

** **

- The tension
in the string is 49 N**T**_{1} - The tension
in the string is**T**_{2} - The tension
in the string is 57 N**T**_{3} - The angle $\theta $ made by
with vertical is 29**T**_{3}^{0}.

The weight of the block A, W_{A} = 40 N

The weight of the block B, W_{B} =50 N

The angle made by T_{1} with vertical is, $\varphi ={35}^{0}$

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:

F_{net }= ma

At equilibrium,

F_{net }= 0

The free body diagram:

At equilibrium,

F_{ynet }= 0

From the figure we can write,

$\backslash \mathrm{begingathered}{W}_{A}={T}_{1}\text{cos}\varphi \phantom{\rule{0ex}{0ex}}{T}_{1}=\frac{{W}_{A}}{\text{cos}\varphi}\phantom{\rule{0ex}{0ex}}=\frac{40}{\text{cos}35\xb0}\phantom{\rule{0ex}{0ex}}=49\text{N}\phantom{\rule{0ex}{0ex}}\backslash \mathrm{endgathered}$

Hence, the tension T_{1} in the string is 49 N

At equilibrium,

${F}_{xnet}=0\phantom{\rule{0ex}{0ex}}{T}_{2}={\text{T}}_{1}\text{sin}\varphi \phantom{\rule{0ex}{0ex}}=49\text{sin}35\xb0\phantom{\rule{0ex}{0ex}}=28\text{N}\phantom{\rule{0ex}{0ex}}$

Hence, the tension T_{2} in the string is* * 28 N

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 **T _{3}** 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 **T _{3}** in the string is 57 N

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

Hence, the angle($\theta $) made by T_{3} with vertical is 29^{0} .

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