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Equilibrium of a Particle - Spring Forces

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Equilibrium of a Particle - Spring 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 

The left-hand image in Figure 1 shows a system of springs in an unstreched state, before a downward force, F, of 800 lb is applied to point P. The right-hand image contains a free-body diagram for the stretch state. 

 

The unstreched spring lengths are in and in and the spring constants are lb/in and lb/in. Since the spring constants (a measure of the stiffness or strength of a spring) are very high, the system is described as stiff. Determine the tension in each spring. 

 

Solve the problem a second time if the spring constants are reduced by a factor of 10.  

Drawing-Canvas 

Figure 1 

 

Solution 

Step 

Result 

The springs in this system creates a system that is more complex than it looks. Since springs are being used instead of rods or ropes, the geometry of the system will change as the force is applied. The general deformation geometry is shown in the right-hand image in Figure 1. The deformation in the springs is given by in the left and right springs, respectively. The angles for the deformed state are and  

 

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

 

HyperlinkImage 

 

The first step in this problem would be to assign values to the relevant variables. Let be the magnitude of the force

 

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 [L][_][1], then press the right arrow to move out of the subscript. 

 

Press [Enter] to evaluate.  

 

 

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800, 8.57, 7.00, 9.56 (1)
 

 

 

As the stiffnesses of the springs decrease, or as the load on the system increases, the lengths of the springs will increase, causing the angles and to increase.  

 

Using the law of cosines, express the angles and in terms of the spring deformations. 

 

To obtain an expression for , first enter the relevant form of the law of cosines for a triangle. 

 

To enter superscript characters, press [Shift][6]. Press the right arrow key to return to the baseline expression. 

 

To enter greek symbols into the equation, use the Greek palette and find the appropriate symbols. 

 

Right-click and select Solve > Isolate Expression for > cos(alpha).With the cursor over the output, de-select, View > Inline Document Output. An equation label will now appear next to the output. 

 

Now repeat the step above for cos( rather than . 

 

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cos(alpha) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(7.00, delta[2]), 2))), `*`(`^`(`+`(8.57, delta[1]), 2)), 91.3936)), `*`(`+`(163.8584, `*`(19.12, `*`(delta[1]))))) (2)
 

 

 

 

cos(beta) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(8.57, delta[1]), 2))), `*`(`^`(`+`(7.00, delta[2]), 2)), 91.3936)), `*`(`+`(133.8400, `*`(19.12, `*`(delta[2]))))) (3)
 

 

In the springs, the tensions and are linearly related to the deformations and by the spring constants, and

 

Press [Enter] to evaluate each expression. 

 

T[1] = `*`(k[1], `*`(delta[1])) (4)
 

 

T[2] = `*`(k[2], `*`(delta[2])) (5)
 

The system is in equilibrium. Therefore, the sum of all the forces acting in the system must be equal to zero.  
 

For this problem it is simpler to resolve the tensile forces into their x- and y-components and have two separate equations of equilibrium. Thus, in component form, we have  

 

 

Drawing-Canvas 

 

 

Enter both the x-component and y-component tensil force equations. 

 

Press [Enter] to evaluate each of the two equations. 

 

`+`(`-`(`*`(T[1], `*`(cos(alpha)))), `*`(T[2], `*`(cos(beta)))) = 0 (6)
 

 

`+`(`*`(T[1], `*`(sin(alpha))), `*`(T[2], `*`(sin(beta))), `-`(800)) = 0 (7)
 

 

There are now six equations for the six unknowns, , , , , and . Therefore, the system of equations is fully determinate.  

 

Enter reasonable starting estimates for and . Try 2000 and 1000 for and respectively.  

 

To solve the system, list the six equations via their equation label. Press [Ctrl][L], then enter the appropriate reference equation number.  

 

Press [Enter].  

 

Right-click and select Solve > Numerically Solve..  

 

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2000, 1000 (8)
 

 

cos(alpha) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(7.00, delta[2]), 2))), `*`(`^`(`+`(8.57, delta[1]), 2)), 91.3936)), `*`(`+`(163.8584, `*`(19.12, `*`(delta[1]))))), cos(beta) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(8.5...
cos(alpha) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(7.00, delta[2]), 2))), `*`(`^`(`+`(8.57, delta[1]), 2)), 91.3936)), `*`(`+`(163.8584, `*`(19.12, `*`(delta[1]))))), cos(beta) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(8.5...
cos(alpha) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(7.00, delta[2]), 2))), `*`(`^`(`+`(8.57, delta[1]), 2)), 91.3936)), `*`(`+`(163.8584, `*`(19.12, `*`(delta[1]))))), cos(beta) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(8.5...
cos(alpha) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(7.00, delta[2]), 2))), `*`(`^`(`+`(8.57, delta[1]), 2)), 91.3936)), `*`(`+`(163.8584, `*`(19.12, `*`(delta[1]))))), cos(beta) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(8.5...
(9)
 

{alpha = -7.000075254, beta = 5.247303265, T[1] = -414.6606875, T[2] = -613.2074874, delta[1] = -.2073303437, delta[2] = -.6132074874}
{alpha = -7.000075254, beta = 5.247303265, T[1] = -414.6606875, T[2] = -613.2074874, delta[1] = -.2073303437, delta[2] = -.6132074874}
{alpha = -7.000075254, beta = 5.247303265, T[1] = -414.6606875, T[2] = -613.2074874, delta[1] = -.2073303437, delta[2] = -.6132074874} 

 

From the computed values of and , it can be seen that the overall geometry of the spring system did not change by much. 

 

To solve the problem using different values for and , redefine the values for these parameters and repeat the above steps to solve the new system of equations. 

 

The system in this case is less rigid and the weight is more equally distributed with both tensions being almost equal. The deformed geometry closely resembles an equilateral triangle. 

 

200, 100 (10)
 

 

 

cos(alpha) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(7.00, delta[2]), 2))), `*`(`^`(`+`(8.57, delta[1]), 2)), 91.3936)), `*`(`+`(163.8584, `*`(19.12, `*`(delta[1]))))), cos(beta) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(8.5...
cos(alpha) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(7.00, delta[2]), 2))), `*`(`^`(`+`(8.57, delta[1]), 2)), 91.3936)), `*`(`+`(163.8584, `*`(19.12, `*`(delta[1]))))), cos(beta) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(8.5...
cos(alpha) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(7.00, delta[2]), 2))), `*`(`^`(`+`(8.57, delta[1]), 2)), 91.3936)), `*`(`+`(163.8584, `*`(19.12, `*`(delta[1]))))), cos(beta) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(8.5...
cos(alpha) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(7.00, delta[2]), 2))), `*`(`^`(`+`(8.57, delta[1]), 2)), 91.3936)), `*`(`+`(163.8584, `*`(19.12, `*`(delta[1]))))), cos(beta) = `/`(`*`(`+`(`-`(`*`(`^`(`+`(8.5...
(11)
 

{alpha = -6.429786219, beta = 4.517864510, T[1] = 164.0953745, T[2] = -839.8095134, delta[1] = .8204768726, delta[2] = -8.398095134}
{alpha = -6.429786219, beta = 4.517864510, T[1] = 164.0953745, T[2] = -839.8095134, delta[1] = .8204768726, delta[2] = -8.398095134} 

 

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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