Forces in equilibrium 1

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Question statement

A load of $W=$ {W} N is suspended from a roof by two cables, A and B.

Cable A makes an angle $\theta_1 =$ {theta_1}° with the roof, while cable B makes an angle $\theta_2 =$ {theta_2}° with the roof.

{geogebra_applet('cbqunudv',[['angle1',theta_1+'°'],['angle2',theta_2+'°'],['W',W]])}

Calculate the magnitude of the tension in cables A and B.

Give any introductory information the student needs.

			Name	Type	Generated Value

base_angle	integer	74
theta_1	integer	68
theta_2	integer	38
W	integer	130

			Name	Type	Generated Value

			T_A	number	107
			T_B	number	50.7

			Name	Type	Generated Value

unit

string

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Generated value: `integer`

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theta_2

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My solution using a force triangle Gap-fill
1. Gap Tension in cable A Engineering Accuracy with units
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2. Gap Tension in cable B Engineering Accuracy with units
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My solution using components Gap-fill
1. Gap Tension in cable A Engineering Accuracy with units
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2. Gap Tension in cable B Engineering Accuracy with units
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Worked Solution

Given values:

$W = \var{W}$ N, $\theta_1 = \var{theta_1}°$, and $\theta_2 = \var{theta_2}°$,

The solution using a force triangle is as follows:

{geogebra_applet('ncuhpcre',[['angle1',theta_2+'°'],['angle2',theta_1+'°']])}

$\begin{align}
\frac{T_A}{sin\ (90^\circ-\theta_2)}&=\frac{W}{sin\ (\theta_1+\theta_2)} & \frac{T_B}{sin\ (90^\circ-\theta_1)}&=\frac{W}{sin\ (\theta_1+\theta_2)}\\\\ T_A&=W\frac{sin\ (90^\circ-\theta_2)}{sin\ (\theta_1+\theta_2)} & T_B&=W\frac{sin\ (90^\circ-\theta_1)}{sin\ (\theta_1+\theta_2)}\\
T_A&=\var{W}\ \frac{sin\ (90^\circ-\var{theta_2}^\circ)}{sin\ (\var{theta_1}^\circ+\var{theta_2}^\circ)} & T_B&=\var{W}\ \frac{sin\ (90^\circ-\var{theta_1}^\circ)}{sin\ (\var{theta_1}^\circ+\var{theta_2})^\circ}\\
T_A&=\var{T_A} \ \var{unit} & T_B&=\var{T_B} \ \var{unit}
\end{align}$

The solution using components is as follows:

{geogebra_applet('rzbj2sxt',[['angle1',theta_1+'°'],['angle2',theta_2+'°'],['W',W]])}

Since the system is in equilibrium, we can say:

$\begin{align}
\Sigma_x &=0 &&& \Sigma_y =0 &\\
T_Bcos\theta_2 -T_Acos\theta_1&=0 & (1) && T_Asin\theta_1 +T_Bsin\theta_2-W=0 && (2)\\
\end{align}$

Solve for $T_A$ and $T_B$ using simultaneous equations.

Rearranging (1)

$\begin{align}T_B&=\frac{T_Acos\theta_1}{cos\theta_2} =T_A\frac{cos\theta_1}{cos\theta_2}& (3) \\ \end{align}$

Substitute (3) into (2)

$\begin{align}
T_Asin\theta_1+T_A\frac{cos\theta_1}{cos\theta_2}sin\theta_2 -W &=0\\
T_A \left(sin\theta_1+\frac{cos\theta_1}{cos\theta_2}sin\theta_2\right)-W&=0\\
T_A&=\frac{W}{ \left(sin\theta_1+\frac{cos\theta_1}{cos\theta_2}sin\theta_2\right)}\\
T_A&=\frac{\var{W}}{ \left(sin\ \var{theta_1}^\circ+\frac{cos\ \var{theta_1}^\circ}{cos\ \var{theta_2}^\circ}sin\ \var{theta_2}^\circ\right)}\\
T_A&=\var{T_A} \ \var{unit}
\end{align}$

Substitute the value of $T_A$ into (3) to find $T_B$

$\begin{align}
T_B&=T_A\frac{cos\theta_1}{cos\theta_2}\\
T_B&=\var{T_A}\ \frac{cos\ \var{theta_1}^\circ}{cos\ \var{theta_2}^\circ}\\
T_B&=\var{T_B} \ \var{unit}
\end{align}$

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This question is used in the following exam:

Chapter 1 Exercises by Andrew Burkert in Andrew's workspace.

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Forces in equilibrium 1

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Taxonomy: mathcentre

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Andrew Burkert 1 year, 10 months ago

Andrew Burkert 1 year, 10 months ago

Andrew Burkert 2 years, 1 month ago

Error in variable testing condition

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Generated value: integer

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Worked Solution

Generated value: `integer`