Question

Draw a detailed force body diagram for this problem (Don't have to solve it)!

The system in the figure below is in equilibrium, with the string in the center exactly horizontal. Block A weighs 39.0 N, block B weighs 55.0 N, and angle φ is 34.0˚. What are

Find (a) tension T₁, (b) tension T₂, (c) tension T₃, and (d) angle θ.

Answer 1

T1cos34 T3cose T1sin34 T3sine WA

Consider Joint A

Consider all vertical forces

$\sum F_{y}=0$

$T_{1}cos\phi -W_{A}=0$

$T_{1}cos\phi =W_{A}$

$T_{1}=\frac{W_{A}}{cos\phi }$

$T_{1}=\frac{39.0N}{cos34 }$

(a) ANSWER: ${\color{Red} T_{1}=47.0425N}$

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Consider Joint A

Consider all horizontal forces

$\sum F_{x}=0$

$T_{2}-T_{1}sin\phi =0$

$T_{2}=T_{1}sin\phi$

$T_{2}=47.0425N*sin34$

(b) ANSWER: ${\color{Red} T_{2}=26.306N}$

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Consider Joint B

Consider all horizontal forces

$\sum F_{x}=0$

$T_{3}sin\theta -T_{2}=0$

$T_{3}sin\theta =T_{2}$

${\color{Red} T_{3}sin\theta =26.306N}$--------(1)

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Consider Joint B

Consider all vertical forces

$\sum F_{y}=0$

$T_{3}cos\theta -W_{B}=0$

$T_{3}cos\theta =W_{B}$

${\color{Red} T_{3}cos\theta =55N}$--------(2)

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Divide (1) by (2)

$\frac{T_{3}sin\theta }{T_{3}cos\theta }=\frac{26.306N}{55N}$

$\frac{sin\theta }{cos\theta }=\frac{26.306}{55}$

$tan\theta =\frac{26.306}{55}$

$\theta =tan^{-1}(\frac{26.306}{55})$

(d) ANSWER: ${\color{Red} \theta =25.56^{\circ}}$

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Put value of $\theta$ in (1)

$T_{3}sin25.56 =26.306N$

$T_{3}=\frac{26.306N}{sin25.56 }$

(c) ANSWER: ${\color{Red} T_{3}=60.97N}$

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