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Equilibrium of a Particle - Tension Forces 3

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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 using 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 the Image buttons to watch the videos. 

The steps in the document can be repeated to solve similar problems. 

Problem Statement  

A 13 kg collar at A and a 10 kg collar at B slide on smooth rods as shown in Figure 1. The collars are connected by a 2 m long cable. If the system is in equilibrium, determine the distance from the bottom of each rod to each collar and the tension in the 2 m cable.  

 

(Assume that g = 9.81 m/.) 

Drawing-Canvas 

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  

There are three forces acting on each collar: the normal force exerted by the smooth rods, the gravitational force due to the mass of the collar, and the tensile force exerted by the cable. These three forces are displayed in Figure 2 to the right. 

 

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 system is in equilibrium, the following two equilibrium equations can be obtained. 

 

 

 

 

Drawing-Canvas 

Figure 2 

 

Define , and for the masses of the collars at A and B respectively; for the gravitational constant; and for the length of the cable. 

 

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 [mass][_][a], then press the right arrow to move out of the subscript. 

 

Press [Enter] to evaluate.  

 

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13, 10, 9.81, 2
 

 

 

The weight of a collar is the product of its mass and the gravitational constant.  The associated forces, and , are downward, the direction in which is negative. 

 

Define , as position vectors for the points   

 

Use the Matrix palette. Set the number of rows to three 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. 

 

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

 

Press [Enter] to evaluate.  

 

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Let and be the respective distances of and from the bottom of the rod on which they are located. Position vectors A and B for points A and B are given in terms of unit vectors and that point upward from the base of the respective rod. 

 

The notation for the magnitude of a vector is found in the common symbol palette ( ? ), or by typing two vertical bars from the keyboard. 

 

Press [Ctrl][=] to evaluate the unit vectors inline. 

 

 

 

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The positions A and B are simply the base positions of the rods, plus the unit vectors, multiplied by the respective distances a and b.  

 

= = = =  

Define the unit vector along the cable pointing from to  

 

Press [Ctrl][=] to evaluate the unit vectors inline. 

 

Now, the vector T can be expressed as a product of the tension's magnitude, t, and the newly obtained unit vector. 

=

If is the magnitude of the tension T, then and the equilibrium equations become 

 

 

 

 

Since the normal forces and are orthogonal to the rods containing points and respectively, we have:  

 

 

 

Taking the dot product of the first equilibrium equation with , and the dot product of the second equilibrium equation with results in two equations with the three unknowns  

 

The dot product can be calculated by using the centered dot from the Common Symbols palette. It can also be calculated by using a period

 

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`+`(`*`(`/`(3, 13), `*`(`^`(13, `/`(1, 2)), `*`(`+`(`-`(127.53), `*`(t, `*`(`+`(`*`(`/`(3, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 26), `*`(a, `*`(`^`(13, `/`(1, 2))))))))))))), `-`(`*`(...
`+`(`*`(`/`(3, 13), `*`(`^`(13, `/`(1, 2)), `*`(`+`(`-`(127.53), `*`(t, `*`(`+`(`*`(`/`(3, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 26), `*`(a, `*`(`^`(13, `/`(1, 2))))))))))))), `-`(`*`(...
(1)
 

 

`+`(`*`(`/`(1, 10), `*`(`^`(10, `/`(1, 2)), `*`(t, `*`(`+`(1, `-`(`*`(`/`(1, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))))))))), `*`(`/`(3, 10), `*`(`^`(10, `/`(1, 2)), `*`(`+`(`-`(`*`(t, `*`(`+`(`*`(`/`(3, ...
`+`(`*`(`/`(1, 10), `*`(`^`(10, `/`(1, 2)), `*`(t, `*`(`+`(1, `-`(`*`(`/`(1, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))))))))), `*`(`/`(3, 10), `*`(`^`(10, `/`(1, 2)), `*`(`+`(`-`(`*`(t, `*`(`+`(`*`(`/`(3, ...
(2)
 

 

A third equation is needed. The constraint on the length of the cable provides this equation.  

 

Note, squaring both sides removes the radical, further simplifying the equation. 

 

`+`(`*`(`^`(`+`(2, `-`(`*`(`/`(1, 10), `*`(b, `*`(`^`(10, `/`(1, 2))))))), 2)), `*`(`^`(`+`(`*`(`/`(3, 10), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 13), `*`(a, `*`(`^`(13, `/`(1, 2))))))), 2)...
`+`(`*`(`^`(`+`(2, `-`(`*`(`/`(1, 10), `*`(b, `*`(`^`(10, `/`(1, 2))))))), 2)), `*`(`^`(`+`(`*`(`/`(3, 10), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 13), `*`(a, `*`(`^`(13, `/`(1, 2))))))), 2)...
(3)
 

 

Form a sequence of the three equations and solve.  

 

Use equation labels to form the sequence. Press [Ctrl][L], then enter the appropriate reference equation number. 

 

Right-click the sequence and select Solve > Solve

 

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`+`(`*`(`/`(3, 13), `*`(`^`(13, `/`(1, 2)), `*`(`+`(`-`(127.53), `*`(t, `*`(`+`(`*`(`/`(3, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 26), `*`(a, `*`(`^`(13, `/`(1, 2))))))))))))), `-`(`*`(...
`+`(`*`(`/`(3, 13), `*`(`^`(13, `/`(1, 2)), `*`(`+`(`-`(127.53), `*`(t, `*`(`+`(`*`(`/`(3, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 26), `*`(a, `*`(`^`(13, `/`(1, 2))))))))))))), `-`(`*`(...
`+`(`*`(`/`(3, 13), `*`(`^`(13, `/`(1, 2)), `*`(`+`(`-`(127.53), `*`(t, `*`(`+`(`*`(`/`(3, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 26), `*`(a, `*`(`^`(13, `/`(1, 2))))))))))))), `-`(`*`(...
`+`(`*`(`/`(3, 13), `*`(`^`(13, `/`(1, 2)), `*`(`+`(`-`(127.53), `*`(t, `*`(`+`(`*`(`/`(3, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 26), `*`(a, `*`(`^`(13, `/`(1, 2))))))))))))), `-`(`*`(...
`+`(`*`(`/`(3, 13), `*`(`^`(13, `/`(1, 2)), `*`(`+`(`-`(127.53), `*`(t, `*`(`+`(`*`(`/`(3, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 26), `*`(a, `*`(`^`(13, `/`(1, 2))))))))))))), `-`(`*`(...
`+`(`*`(`/`(3, 13), `*`(`^`(13, `/`(1, 2)), `*`(`+`(`-`(127.53), `*`(t, `*`(`+`(`*`(`/`(3, 20), `*`(b, `*`(`^`(10, `/`(1, 2))))), `-`(`*`(`/`(3, 26), `*`(a, `*`(`^`(13, `/`(1, 2))))))))))))), `-`(`*`(...
(4)
 

 

{a = 4.106809533, b = 3.267767429, t = 306.9416523}, {a = 10.31539557, b = 9.381343212, t = -306.9416523}
{a = 4.106809533, b = 3.267767429, t = 306.9416523}, {a = 10.31539557, b = 9.381343212, t = -306.9416523}
(5)
 

 

Of the two solutions returned, select the one in which , the magnitude of the tension, is positive. This solution also has the property that neither nor exceeds the length of the relevant rod. 

 

Assuming that the correct solution is the first of the two, right-click the solutions and select Select Elements > 1. 

 

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{a = 4.106809533, b = 3.267767429, t = 306.9416523}, {a = 10.31539557, b = 9.381343212, t = -306.9416523}
{a = 4.106809533, b = 3.267767429, t = 306.9416523}, {a = 10.31539557, b = 9.381343212, t = -306.9416523}
(6)
 

 

{a = 4.106809533, b = 3.267767429, t = 306.9416523} (7)
 

 

 

The normal forces and can be obtained from the equilibrium equations and substitution of the values of

 

Use the template from the Expression palette to evaluate the expression at a point. Use an equation label to reference the desired point to evaluate at. 

 

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(8)
 

 

 

Compute the magnitudes of the forces and , namely and respectively. 

 

 

For the output of each magnitude, right-click and select Simplify. 

 

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`*`(`^`(`+`(`*`(`^`(`+`(`-`(306.9416523), `*`(50.15069668, `*`(`^`(10, `/`(1, 2))))), 2)), `*`(`^`(`+`(127.53, `-`(`*`(150.4520901, `*`(`^`(10, `/`(1, 2))))), `*`(145.4481812, `*`(`^`(13, `/`(1, 2))))...
`*`(`^`(`+`(`*`(`^`(`+`(`-`(306.9416523), `*`(50.15069668, `*`(`^`(10, `/`(1, 2))))), 2)), `*`(`^`(`+`(127.53, `-`(`*`(150.4520901, `*`(`^`(10, `/`(1, 2))))), `*`(145.4481812, `*`(`^`(13, `/`(1, 2))))...
`*`(`^`(`+`(`*`(`^`(`+`(`-`(306.9416523), `*`(50.15069668, `*`(`^`(10, `/`(1, 2))))), 2)), `*`(`^`(`+`(127.53, `-`(`*`(150.4520901, `*`(`^`(10, `/`(1, 2))))), `*`(145.4481812, `*`(`^`(13, `/`(1, 2))))...
(9)
 

 

350.5504875 (10)
 

 

`*`(`^`(`+`(`*`(`^`(`+`(306.9416523, `-`(`*`(50.15069668, `*`(`^`(10, `/`(1, 2)))))), 2)), `*`(`^`(`+`(98.10, `*`(150.4520901, `*`(`^`(10, `/`(1, 2)))), `-`(`*`(145.4481812, `*`(`^`(13, `/`(1, 2))))))...
`*`(`^`(`+`(`*`(`^`(`+`(306.9416523, `-`(`*`(50.15069668, `*`(`^`(10, `/`(1, 2)))))), 2)), `*`(`^`(`+`(98.10, `*`(150.4520901, `*`(`^`(10, `/`(1, 2)))), `-`(`*`(145.4481812, `*`(`^`(13, `/`(1, 2))))))...
`*`(`^`(`+`(`*`(`^`(`+`(306.9416523, `-`(`*`(50.15069668, `*`(`^`(10, `/`(1, 2)))))), 2)), `*`(`^`(`+`(98.10, `*`(150.4520901, `*`(`^`(10, `/`(1, 2)))), `-`(`*`(145.4481812, `*`(`^`(13, `/`(1, 2))))))...
(11)
 

 

307.0695986 (12)
 

 

Legal Notice: The copyright for this application is owned by Maplesoft. The application is intended to demonstrate the use of Maple to solve a particular problem. It has been made available for product evaluation purposes only and may not be used in any other context without the express permission of Maplesoft.   

 

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