# Simple Form Truss—Classroom Set

### Materials Included In Kit

Binder clips, 4
Meter stick, half, with hole
String, thin, 2 m
Support clamp with bracket, screw and bolt

(for each setup)
C-clamp, 3" (optional, recommended)
Hooked mass, 500-g
Protractor
Ruler or meter stick
Scissors
Spring scale, 1000-g
Support stand

### Prelab Preparation

Fasten the screw through the hole in the half meter stick and through the bracket on the support stand clamp. Tighten the bolt assembly until the meter stick does not twist but can still be raised and lowered easily.

{12662_Preparation_Figure_11}

### Safety Precautions

The materials in this lab are considered safe. Please follow normal laboratory safety guidelines.

### Disposal

The materials should be saved for future classes.

### Lab Hints

• Enough materials are provided in this kit for eight student groups. Additional kits should be purchased depending on the classroom goals. The experiment can be reasonably completed in two 50-minute class periods.
• Assume the mass of the boom and binding clips is negligible.
• If 500-g masses are not available, any small, heavy object can be used. A small bag of sand is one low-cost option. The design of the experiment may need to be changed slightly (e.g., having to hang the mass over the edge of a table if it is too large)
• To limit the metal fatigue in the binding clips, remove the clips from the meter stick before storing. Two additional clips have been provided if binding clips become too weak to securely clamp onto the meter stick. The binding clips can also be purchased at a local office supply.

### Teacher Tips

• This lab can also be inquiry-based. Provide the students only with the necessary task and have them develop their own procedure, identifing the variables, developing a data table and analyzing their results.
• In order to determine the length of string, use simple right-triangle geometry. Assume that the attachment point of the string is directly above the pivot point.
sin θ = opposite side/hypotenuse
tan θ = opposite side/adjacent side
{12662_Tips_Figure_12}

### Science & Engineering Practices

Developing and using models
Planning and carrying out investigations
Analyzing and interpreting data
Using mathematics and computational thinking
Constructing explanations and designing solutions
Obtaining, evaluation, and communicating information

### Disciplinary Core Ideas

MS-ETS1.A: Defining and Delimiting Engineering Problems
MS-ETS1.B: Developing Possible Solutions
MS-ETS1.C: Optimizing the Design Solution
HS-ETS1.A: Defining and Delimiting Engineering Problems
HS-ETS1.B: Developing Possible Solutions
HS-ETS1.C: Optimizing the Design Solution

### Crosscutting Concepts

Systems and system models
Stability and change
Structure and function

### Performance Expectations

MS-PS1-2: Analyze and interpret data on the properties of substances before and after the substances interact to determine if a chemical reaction has occurred.
HS-PS1-2: Construct and revise an explanation for the outcome of a simple chemical reaction based on the outermost electron states of atoms, trends in the periodic table, and knowledge of the patterns of chemical properties.
HS-PS1-5: Apply scientific principles and evidence to provide an explanation about the effects of changing the temperature or concentration of the reacting particles on the rate at which a reaction occurs.
HS-PS1-6: Refine the design of a chemical system by specifying a change in conditions that would produce increased amounts of products at equilibrium.
HS-PS1-7: Use mathematical representations to support the claim that atoms, and therefore mass, are conserved during a chemical reaction.

1. Use simple right-hand triangle geometry to determine the length of string (l) needed to support a 50-cm-long boom when the string is held at a 45° angle and the boom is held at 30° with respect to the ground. Assume the string is attached to the wall directly above the pivot point of the boom. Hint: Break the triangle into two separate right-hand triangles with a common side. Then solve for the common side (see Figure 3).

Adjacent side = (50 cm) x (cos 30°) = 43.3 cm
l = (43.3 cm)/(cos 45°) = 61.2 cm

2. Using your solution to Prelab Question 1 as a guide, calculate the string lengths for the different string positions, string angles, and boom angles given in the data table. Assume the string “attachment point” is directly above the pivot point of the boom. Record these values in the data table.

See Data Tables. Note: The string lengths were calculated assuming the boom is 50 cm long. In this setup, however, the boom is actually 49 cm long because the pivot point is at the 1-cm mark, not the zero mark. So, the value of the positions of the string and mass will be one higher than their actual distance from the pivot. Some students may recognize this condition. Whether students use the value in the data table or the actual distance, the string length will not vary considerably (by at most 2) and will not affect the students’ conclusions as long as they are consistant.

3. In Figure 4, if a mass was hanging at the 40-cm point on the boom, how high above the ground would the mass attachment point be? Assume the pivot point is 25 cm above the ground and the mass attachment point is at the mid-line of the boom where the mass is hanging. Hint: Use right-angle triangle geometry.

Opposite side = (40 cm) x (sin 30°) = 20 cm
Height above the ground = 25 cm + 20 cm = 45 cm

### Sample Data

{12662_Data_Table_1}
{12662_Data_Table_2}

1. Which boom setup completed the objectives in the best possible manner (i.e., the sign is hanging as far away from the wall as possible using the least amount of string)? How much did it cost for the cable?

From the experimental results, the best setup is Test 11 in which the mass is positioned at 25 cm, the string attachment is at 40 cm and the angle of the boom is 45º. This produced a string tension of 200 g. The position of the mass (“sign”) is at 18 cm from the ring stand (wall) and is 38 cm high (18 cm + 20 cm initial height). The amount of string that was needed was 40 cm, resulting in a material cost of \$200.
Test number 6 does not fit the criteria because the sign will hang higher than 40 cm. It will hang from a height of 55 cm above the tabletop, which is “too dangerous.”
Test 30 is another possibility. The sign will only hang 12.5 cm from the “wall,” but the material cost will only be \$90.

2. In general, in order to provide the strongest support, where is the best place to hang the mass in relation to the string attachment?

The best place to position the mass in relation to the string is to place the mass as close to the wall as possible and attach the string as far from the wall as possible.

3. How does the angle of the boom affect the tension in the string?

As the angle of the boom increases, the tension in the string decreases (all other variables being equal).

4. Compare the force measurements for when the string and mass are attached at the same position.

The force measurements for when the string and mass are attached at the same position are very similar. It does not matter where on the boom the string and mass are attached, so long as they are attached at the same point. Only the angle of the boom and the angle of the string affected the tension in the string.

5. Which boom would produce the most tension in the string? Why? Assume the mass in each figure is the same.

The boom shown in Figure 8c would result in the highest string tension. This is because the angle of the string is the only important variable when the mass and string are attached at the same location. The smaller the angle of the string leads to a larger tension in the string. Therefore, Figure 8c would show the highest tension in the string.

6. (Optional) For some experiments, the tension in the string is greater than the actual weight of the object. Why is the tension in the string larger than the weight of the object?

The “extra” weight comes from the torque about the pivot point. Torque is equal to the perpendicular length multiplied by the weight of the object. The vertical component of the tension in the string must equal the weight of the object, otherwise the forces would not be balanced and the boom would rotate downward. Since the string is at an angle, the vertical tension is equal to the weight of the mass, but is less than the total tension in the string. The total tension in the string is a combination of both vertical and horizontal tension component: Tt = √[(Tx)2 + (Ty)2].

7. (Optional) For other experiments, the tension in the string is less than the actual weight of the object. Why does it take less cable strength to support the object? Where is the additional lifting force coming from?

The cable strength is reduced because the pivot point attached to the wall provides more upward force to balance the weight of the object. The pivot point provides more force when the string is attached at a further distance away than the mass is from the pivot point. It also provides more upward force when the boom is at an upward angle with respect to the horizontal.

8. (Optional) For the “optimal” boom setup in Question 1, calculate the horizontal and vertical forces exerted on the truss by the ring stand at the pivot point. Hint: Units need to be in meters and kilograms.

T = (0.5 kg)(9.81 m/s2)(0.25 m)sin (90° – 45°)/(0.40 m)sin (45° + 45°) = 2.2 N
Wx = (2.2 N) cos (45°) = 1.6 N
Wy = (0.5 kg)(9.81 m/s2) – (2.2 N) sin (45°) = 3.3 N

### Discussion

When a lever arm is made to pivot around a fulcrum, a rotational force must act on the lever arm. This rotational force is called torque. A torque is actually the perpendicular force (F) multiplied by the distance (D) between where the force is applied and the position of the pivot point (Equation 1).

{12662_Discussion_Equation_1}
The “sin θ” term is included to make sure only the perpendicular components of the force and lever arm distance are multiplied together. When the angle between the force and the lever arm is perpendicular (90°) then the sin θ term becomes 1 (sin 90° = 1).

The conditions for static equilibrium occur when the net force acting on the rigid body and the net torque about any point on the rigid object are both equal to zero (Equations 2 and 3).
{12662_Discussion_Equation_2}
{12662_Discussion_Equation_3}
The free-body diagram for a general simple form truss is shown in Figure 13.
{12662_Discussion_Figure_13}
The equations for the torque produced by the hanging mass (τm) and the cable (τc) are shown in Equations 4 and 5.
{12662_Discussion_Equation_4}
{12662_Discussion_Equation_5}
Where mg is the weight of the mass (g = 9.81 m/s2), D is the distance the mass hangs from the pivot point; T is the tension in the cable; L is the distance of the cable attachment location from the pivot point. β is equal to (90 – θ). The “positive” direction is chosen to be in the counterclockwise direction. Therefore, the net torque equation is equal to
{12662_Discussion_Equation_6}
Rearranging to solve for the tension in the cable
{12662_Discussion_Equation_7}
Equation 7 is the general equation for the tension in the string of a simple form truss. If the simple form truss is held horizontally and the mass and cable are at the same location on the boom then Equation 7 simplifies to
{12662_Discussion_Equation_8}
There is more to the simple form truss than simply the tension in the string. The wall at the pivot point also applies a force in order to maintain static equilibrium. The net force must also be equal to zero. The force provided by the wall at the pivot point can be determined by separating the force into the x- and y-components.
{12662_Discussion_Equation_9}
{12662_Discussion_Equation_10}
Where Wx and Wy represent the x- and y-components of the force applied by the wall.
{12662_Discussion_Equation_11}
{12662_Discussion_Equation_12}
As an example, suppose a 1000-g mass is hanging at a distance of 25 cm from the pivot point of the boom and that the boom is at a 30° angle. The boom is supported by a cable at a 45° angle and attached 40 cm from the pivot point of the boom. What is the tension in the string and the force applied by the wall?

T = (1 kg)(9.81 m/s2)(0.25 m) sin (90° – 30°)/(0.40 m) sin (45° + 30°) = 5.5 N
Wx = (5.5 N) cos (45°) = 3.9 N
Wy = (1 kg)(9.81 m/s2) – (5.5 N) sin (45°) = 5.9 N

The total force exerted by the wall at the pivot point is equal to
{13047_Discussion_Equation_13}
It should be noted that the forces provided by the wall and the tension in the string are greater than the overall weight of the hanging mass—7.1 N + 5.5 N > 9.8 N.

# Simple Form Truss

### Introduction

The ability to make strong, rigid structures has been important ever since the first buildings were constructed many thousands of years ago. In modern times, structural strength is even more important with the construction of complex bridges and skyscrapers. All these structures have the same physical property in common—they are all in static equilibrium. This laboratory activity introduces the concept of static equilibrium. Your task, as an engineer, will be to hang a “sign” over a sidewalk for the lowest material cost.

### Concepts

• Trusses and boom supports
• Torque
• Second- and third-class lever arms
• Static equilibrium of a rigid body

### Background

Static equilibrium occurs when all the forces acting on a structure are in perfect balance. That is, there is no linear or rotational movement. If a building or bridge is not in static equilibrium, the unbalanced forces, the most significant being from the force due to gravity, will eventually cause the structure to fall.

The ability to maintain static equilibrium becomes more difficult when an object must be supported from above instead of below. Long suspension bridges are generally supported by wire cables that attach to the tops of the supporting bases (see Figure 1). Overhanging signs and cranes are similar to suspension bridges and rely heavily on strong cables, a supporting lever, also called a boom or truss, and cable attachments. In order to save costs, engineers attempt to limit the amount of material used to support a structure, while still maintaining a high level of strength so that the structure stays in static equilibrium for many years to come.

{12662_Background_Figure_1}
A simple truss is a supporting structure consisting of a lever arm (boom) and a supporting cable. A simple truss can act as either a Class II lever or a Class III lever, depending on where the supporting cable is in relation to the supported load (see Figure 2). The fulcrum of the truss is the pivot point where it is connected to the supporting wall.
{12662_Background_Figure_2}

### Experiment Overview

Laboratory Objective Guidelines
The boom used to hang the sign must be attached to the wall of the building at a specific location. The sign should be hung as far away from the building as possible in order to maximize its visibility and marketing potential. However, the sign can not be hung above a certain height for safety reasons. Also, in order to keep the costs of the project down and still maintain building code, the “cable” material used can only support half the weight of the sign. If the “cable” attempts to support more than half the weight of the sign (250 g; 2.45 N), the “cable” will snap. In order to accomplish this task, experiment with the position of the sign on the boom, the angle of the boom, and the position of the supporting rope. Lab Hint: Hang the mass and attach the string at 10 cm, 25 cm and 40‑cm on the meter stick, and use boom angles of 30°, 45° and 60°.

• Sign mass: 500 g
• Boom attachment height on wall: 20 cm
• Maximum height of sign attachment point: 40 cm
• Maximum supporting weight of cable: 250 g
• Cost of supporting cable: \$5/cm

### Materials

Binding clips, 2
C-clamp, 3" (optional, but recommended)
Hooked mass, 500-g
Meter stick, half, with support stand clamp (Simple Form Truss)
Protractor
Ruler or metric stick
Scissors
Spring scale, 1000-g
String
Support clamp with bracket
Support stand

### Prelab Questions

1. Use simple right-hand triangle geometry to determine the length of string (l) needed to support a 50-cm-long boom when the string is held at a 45° angle and the boom is held at 30° with respect to the ground. Assume the string is attached to the wall directly above the pivot point of the boom. Hint: Break the triangle into two separate right-hand triangles with a common side. Then solve for the common side (see Figure 3).
{12662_PreLab_Figure_3}
2. Using your solution to Prelab Question 1 as a guide, calculate the string lengths for the different string positions, string angles, and boom angles given in the data table. Assume the string “attachment point” is directly above the pivot point of the boom. Record these values in the data table.
3. In Figure 4, if a mass was hanging at the 40-cm point on the boom, how high above the ground would the mass attachment point be? Assume the pivot point is 25 cm above the ground and the mass attachment point is at the mid-line of the boom where the mass is hanging. Hint: Use right-angle triangle geometry.
{12662_PreLab_Figure_4}

### Safety Precautions

The materials in this lab are considered safe. Please follow normal laboratory safety guidelines.

### Procedure

1. Obtain the Simple Form Truss, two binding clips, support stand, 500-g hooked mass, string, scissors, protractor, ruler and 1000-g spring scale.
2. Use scissors to cut two pieces of string—one approximately 50 cm and the other approximately 15 cm.
3. Tie one end of the 50-cm string to the spring-scale handle.
{12662_Procedure_Figure_5}
{12662_Procedure_Figure_6}
4. The setup for the Simple Form Truss is shown in Figures 5 and 6.

Note 1: This is only a general setup. The angle of the boom, string, and the position of the mass will vary throughout the experiment depending on the necessary tests.
Note 2: The spring-scale scale should face the measurer.
Note 3: Tie the 15-cm string around the binder clip to form a loop that is just large enough to allow the string to slide along the meter stick (see Figure 7). The hooked mass will hang from the string and the binder clip will prevent the string from slipping. The spring-scale binder clip should be clipped on the bottom edge of the meter stick. This will prevent the clips from being pulled off the meter stick.

{12662_Procedure_Figure_7}
Note 4: (Optional) Clamp the support stand to the tabletop using a C-clamp.
Note 5: Figure 6 shows how to use the mass binder clip when the mass and string are attached at the same position on the meter stick. Only one binder clip is used for this setup.
5. Lab partner responsibilities:

Lab partner 1: Hold the base of the support stand with one hand and pull on the string with your other hand to lift the boom to the proper angle as well as keep the string at the proper angle. (If a C-clamp is used, the support stand should be secure and it will not be necessary to hold it down with one hand.)
Lab partner 2: Measure and record the angle of the string and boom with the protractor and the force measurement from the spring scale.

6. Follow steps 7–10 to collect spring-scale force data for each particular test given in the data table. When the mass and string are connected at the same position, set up the boom as shown in Figure 6. All other tests will be set up similar to Figure 5. Hang the mass from the string, making sure the mass clip is secured to the meter stick and does not slip. Both clips should be centered around the appropriate mark on the meter stick (see Figure 8).
{12662_Procedure_Figure_8}
7. With the ring stand firmly held down with one hand, slowly pull on the string toward the ring stand rod in order to raise the boom to the appropriate angle needed for the specific test. Pull the string in the same plane as the boom in order to prevent the boom from spinning around the support stand rod. Note: It may be easier to raise the boom to the proper angle with one hand first and then pull on the spring scale until there is enough tension to support the boom at the desired angle.
8. Slowly raise or lower the string angle, while maintaining the necessary boom angle, until the string and boom are at the appropriate angle according to the values in the data table. Use a protractor to measure the approximate angles of the string (A) and of the boom (B) (see Figure 9). The angles should be within ±3° of the value in the data table for the specific test. All angles should be measured with respect to the horizontal.
{12662_Procedure_Figure_9}
9. When the boom and string are at the proper angles, measure and record the force indicated on the spring scale. Note: Depending on the mass used, some tests may result in a force greater than 1000 g. If this occurs, record >1000 g in the data table.
10. If the positions of the mass and string do not change, it is typically easier and faster to continue adjusting the string and boom angles to those indicated in the data table, until all the measurements are recorded for a specific set of mass and string positions. When all the data has been collected for the particular mass and string positions, slowly lower the mass to the tabletop.
11. Repeat steps 6–10 for each test in the data table, using the appropriate clip positions, boom angle and string angle for each test. Measure and record the force on the spring scale when the boom and string angles are at the appropriate values (within ±3°). Note: It is easier to move the mass clip and looped string when the mass is removed from the string.
12. Consult your instructor for appropriate storage procedures.

### Student Worksheet PDF

12662_Student1.pdf

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