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Equilibrium of a Particle - Tension Forces 2

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Equilibrium of a Particle - Tension Forces 

? Maplesoft, a division of Waterloo Maple Inc., 2008 

 

Introduction 

This application is one of a collection of educational engineering examples with Maple. These applications use Clickable Engineering? methods to solve problems interactively. Steps are given at every stage of the solution, and many are illustrated using short video clips.  

Click on theImage buttons to watch the videos. 

This application is reusable. Modify the problem, then click the !!! button on the toolbar to re-execute the document to solve the new problem. 

 

Problem Statement 

As per Figure 1, two collars and of masses 5 kg and 9 kg respectively, are connected by a cable and hang on a vertical frame composed of two smooth rods with the vertex angle shown.  

 

Determine the equilibrium values for T, the tension in the cable, and , the angle the cable makes with the horizontal. 

 

(Assume that g = 9.81 m/.) 

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Figure 1 

 

 

Solution 

Step 

Result 

To perform matrix computations, load the Student Linear Algebra package. 

 

Tools > Load package > Student Linear Algebra  

 

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Loading Student:-LinearAlgebra  

 

Treat collars P and Q as particles and construct their free-body diagrams.  

 

Diagrams can be drawn by using the Canvas feature in Maple. To insert a Canvas into a worksheet, apply, Insert > Canvas

 

 

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Since the entire system is in equilibrium, the sum of forces acting on P or must equal zero. From the two free-body diagrams shown in Figure 2, obtain the two equilibrium equations: 

 

 

 

 

 

Drawing-Canvas 

Figure 2 

 

In each case, a standard -coordinate system (with basis vectors i and j) is operative.

Define the gravitational constant , and the masses of the collars and .  

 

Use the assignment operator (a colon followed by an equal sign) to define the following variables. 

 

For subscript notation, use the underscore ( _ ) to move the cursor to the subscript position, and the right arrow (→) to move back to the baseline. For example, to enter , type [m][_][p], then press the right arrow to move out of the subscript. 

 

Press [Enter] to evaluate.  

 

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9.81, 5, 9 (3.1)
 

 

With the weight given by , obtain the weights of the two collars, and  

 

 

49.05 (3.2)
 

 

88.29 (3.3)
 

Let be a unit vector in the direction of T, and let be the magnitude of T.   

 

Let and be unit vectors respectively in the directions of the normal forces acting on and .   

 

Let and be the respective magnitude of these normal forces. Thus, we introduce three unknowns, namely, , and

 

Define and  

 

Use the Matrix palette. Set the number of rows to two and the number of columns to one and then press the Insert Vector[column] button. You could also use the Choose button and drag the mouse to select the matrix size. 

 

Obtain θ from the Greek palette.  

 

Fill in an element, then press [Tab] to move to the next placeholder.  

 

Press [Enter] to evaluate.  

 

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(3.4)
 

 

In addition to the tension , enter the vectors , , , and .  

 

Use the Matrix palette to enter the vectors. 

 

Press [Enter] to evaluate.  

 

 


 

The two equilibrium equations represent four equations in the four unknows .  To obtain the first two of these equations, enter the left-hand side of the first equation.  

 

 

Right-click and select Conversions > To List then again Conversions > To Set

 

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(3.5)
 

 

[`+`(`-`(`*`(`/`(1, 2), `*`(n[P], `*`(`^`(2, `/`(1, 2)))))), `*`(t, `*`(cos(theta)))), `+`(`-`(49.05), `*`(`/`(1, 2), `*`(n[P], `*`(`^`(2, `/`(1, 2))))), `*`(t, `*`(sin(theta))))] (3.6)
 

 

{`+`(`-`(`*`(`/`(1, 2), `*`(n[P], `*`(`^`(2, `/`(1, 2)))))), `*`(t, `*`(cos(theta)))), `+`(`-`(49.05), `*`(`/`(1, 2), `*`(n[P], `*`(`^`(2, `/`(1, 2))))), `*`(t, `*`(sin(theta))))} (3.7)
 

 

The remaining two equations are obtained likewise from the second equilibrium equation. 

 

Perform the same operation as the step above. 

 

 

(3.8)
 

 

[`+`(`*`(`/`(1, 2), `*`(n[Q], `*`(`^`(3, `/`(1, 2))))), `-`(`*`(t, `*`(cos(theta))))), `+`(`-`(88.29), `*`(`/`(1, 2), `*`(n[Q])), `-`(`*`(t, `*`(sin(theta)))))] (3.9)
 

 

{`+`(`*`(`/`(1, 2), `*`(n[Q], `*`(`^`(3, `/`(1, 2))))), `-`(`*`(t, `*`(cos(theta))))), `+`(`-`(88.29), `*`(`/`(1, 2), `*`(n[Q])), `-`(`*`(t, `*`(sin(theta)))))} (3.10)
 

 

A set of all four equations is obtained by forming a set union of the two subsets.  

 

Solve this set. 

 

Refer to the Common Symbols palette to obtain the union operator, .  

 

Reference the subsets by their equation labels. Press [Ctrl][L], then enter the appropriate reference equation number. 

 

Right-click the equation and select Solve > Solve

 

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There are two solutions; the meaningful one gives , the magnitude of the tension, as positive. The corresponding angle is negative, indicating that the cable tilts downward from to

 

{`+`(`-`(`*`(`/`(1, 2), `*`(n[P], `*`(`^`(2, `/`(1, 2)))))), `*`(t, `*`(cos(theta)))), `+`(`*`(`/`(1, 2), `*`(n[Q], `*`(`^`(3, `/`(1, 2))))), `-`(`*`(t, `*`(cos(theta))))), `+`(`-`(88.29), `*`(`/`(1, ...
{`+`(`-`(`*`(`/`(1, 2), `*`(n[P], `*`(`^`(2, `/`(1, 2)))))), `*`(t, `*`(cos(theta)))), `+`(`*`(`/`(1, 2), `*`(n[Q], `*`(`^`(3, `/`(1, 2))))), `-`(`*`(t, `*`(cos(theta))))), `+`(`-`(88.29), `*`(`/`(1, ...
(3.11)
 

 

{t = 95.00906838, theta = -.4117056941, n[P] = 123.1356753, n[Q] = 100.5398579}, {t = -95.00906838, theta = 2.729886959, n[P] = 123.1356753, n[Q] = 100.5398579}
{t = 95.00906838, theta = -.4117056941, n[P] = 123.1356753, n[Q] = 100.5398579}, {t = -95.00906838, theta = 2.729886959, n[P] = 123.1356753, n[Q] = 100.5398579}
{t = 95.00906838, theta = -.4117056941, n[P] = 123.1356753, n[Q] = 100.5398579}, {t = -95.00906838, theta = 2.729886959, n[P] = 123.1356753, n[Q] = 100.5398579}
(3.12)
 

 

 

 

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