Materials Included In Kit

Additional Materials Required

Prelab Preparation

Fasten the bolt 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 (see Figure 11).

Safety Precautions

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

Disposal

All materials may be saved for future use.

Lab Hints

• This 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., 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 store.

Teacher Tips

• This lab is inquiry-based. Provide the students only with the necessary task and have them develop their own procedure, identifying the variables, developing a data table, and analyzing their results.
• Instruct students to slowly raise or lower the string angle, while keeping the boom parallel to the ground, until the string is at the appropriate angle according to the value in the data table. Use a protractor to measure the approximate angle of the string (see Figure 12). All angles should be measured with respect to the horizontal.
  {13792_Tips_Figure_12}
• In order to determine the length of string, use simple right-triangle geometry (see Figure 13). Assume the attachment point of the string is directly above the pivot point.
  {13792_Tips_Figure_13}

  sinθ = opposite side/hypotenuse
  cosθ = adjacent side/hypotenuse
  tanθ = opposite side/adjacent side

Further Extensions

Opportunity for Inquiry

Use the same outline to determine the strength of a cable needed to secure a sign from the force of wind gusts, given the area of the sign and the force of the wind.

Alignment to the Curriculum for AP^® Physics 1

Enduring Understandings and Essential Knowledge
A force exerted on an object can cause a torque on that object. (3F)
3F1: Only the force component perpendicular to the line connecting the axis of rotation and the point of application of the force results in a torque about that axis.

Learning Objectives
3F1.1: The student is able to use representations of the relationship between force and torque.
3F1.2: The student is able to compare the torques on an object caused by various forces.
3F1.3: The student is able to estimate the torque on an object caused by various forces in comparison to other situations.
3F1.4: The student is able to design an experiment and analyze data testing a question about torques in a balanced rigid system.
3F1.5: The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction).

Science Practices
1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively.
2.2 The student can justify the selection of a mathematical routine to solve problems.
4.1 The student can justify the selection of the kind of data needed to answer a particular scientific question.
4.2 The student can design a plan for collecting data to answer a particular scientific question.
5.1 The student can analyze data to identify patterns or relationships.
6.1 The student can justify claims with evidence.

Answers to Prelab Questions

1. Use simple right-hand triangle geometry (see Figure 4) to determine the length of string (l) needed to support a 40-cmlong boom when the string is held at a 30° angle and the boom is held parallel with respect to the ground. Assume the string is attached to the wall directly above the pivot point of the boom.

   l = (40 cm)/(cos30°) = 46.2 cm

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

   See sample 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 2 at most) and will not affect the students’ conclusions as long as they are consistent.

Sample Data

Answers to Questions

1. With the 500-g mass and the string attachment points fixed at given distances, does the force needed to maintain equilibrium decrease or increase when the string angle is increased? Explain.

   Decrease. For a given force, as the angle increases, the normal component of force also increases. To retain the same normal component of force, request a smaller overall force at this increased angle.

2. If the distance from the pivot point is increased for the 500-g mass in Question 1 and the string is held at a fixed angle, does the force needed to maintain equilibrium increase? Explain.

   t of torque also increases. The normal component of the string force, and therefore the force on the string, must also increase to balance the increase in torque.

3. Refer to Question 1. If the string attachment point is moved closer to the pivot point and the string angle is fixed, does the force needed to maintain equilibrium increase or decrease? Explain.

   Increase. For a given torque, as the distance from the pivot point decreases for the string, the amount of force must increase to balance the decrease in distance.

• See Sample Data.
• Which boom setup completed the objectives in the best possible manner (i.e., sign hangs 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 17 in which the mass is positioned at 10 cm, and the string attachment is at 25 cm. This produced a string tension of 250 g. The position of the mass (“sign”) is at 10 cm from the ring stand (wall). The amount of string that was needed was 35 cm, resulting in a material cost of $165. This is the least expensive design.

• 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.

• How does the angle of the string affect the tension in the string?

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

• 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 string affected the tension in the string.

• Which boom in Figure 8 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. A smaller string angle leads to a larger tension in the string. Therefore, Figure 8c would show the highest tension in the string.

1. A 20-meter bar has a fixed rotational point at its leftmost edge. A series of forces are exerted on the bar at various locations:
   • A 48 N force, at 10 meters, acting at a 45° angle above the bar
   • A 4 N force, at 12 meters, acting at a 90° angle above the bar
   • An 88 N force, at 15 meters, acting at a 145° angle above the bar>
   • A 60 N force, at 8 meters, acting at a 45° angle below the bar
   • A 100 N force, at 10 meters, acting at a 90° angle below the bar

     Draw the free body diagram and calculate the resultant torque, if any, on the 20-meter bar.
     Σ = [48 N sin(45°) x 10 m] + [4 N x 12 m] + [88 N x sin(35°) x 15 m] – [60 N x sin(45°) x 8 m] – [100 N x 10 m]
     Σ = [339 N • m] + [48 N • m] + [757 N • m] – [339 N • m] – [1000 N • m]
     Σ = –195 N

     {13792_Answers_Figure_14}
2. A truck moves across a bridge. The truck has a mass of 1700 kg. The bridge is 1200 meters long, with 2 piers, each pier located 400 meters from the ends. At the point where the truck has travelled 700 meters across the bridge, what are the forces on each pier? Draw the free-body diagram.

   Since all parts of the bridge are in static equilibrium, two conditions apply:

   {13792_Answers_Figure_15}
   1. The sum of all new forces is zero; F_A + F_B – F_truck = 0
   2. The sum of all new torques is zero; T_A = T_B = T_truck = 0

   T = r x F = (r)(f)(sinθ). All forces are acting at 90° to the bridge, so sin90° = 1. Therefore, T = (r)(f).

   {13792_Answers_Equation_4}
3. 40-gram mass is supported by two ropes. The angles θ₁ and θ₂ are 60° and 45°, respectively. Find the values for T₁ and T₂.
   {13792_Answers_Figure_16}

   At point A:
   ΣF = 0:
   T_m + (–T₁) + (–T₂) = 0
   T_m = T₁ + T₂ and T_m = mg
   Breaking these forces into their x and y components:
   T_m = T_mx + T_m; since T_mx = 0, T_m = T_my
   T₁ = T_1x + T_1y
   T₂ = T_2x + T_2y
   Therefore, T_my = T_1x + T_1y + T_2x + T_2y
   The y components are additive, as are the x components:
   T_my = T_1y + T_2y
   T_mx = 0 = –T_1x + T_2x or T_1x = T_2x
   Using trigonometry:
   T_1x = (cos60°)T₁
   T_2x = (cos45°)T₂
   T₁ = [(cos45°)/(cos60°)]T₂ = 1.4 T₂
   T_my = T_1y + T_2y = (sin60°)T₁ + (sin45°)T₂ = (sin60°)1.4T₂ + (sin45°)T₂
   T_my = [(sin60°)1.4 + (sin45°)]T₂
   T_my = (1.21 + 0.707)T₂ = (1.92)T₂
   T₂ = T_my /1.92 = mg/1.92 = (0.040 kg)(9.81 m /sec²)/1.92 = 0.203 N
   T₁ = (1.4)T₂ = 0.287 N

Introduction

The ability to make strong, rigid structures has been important ever since buildings were first 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

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

Background

If a force, applied to an object, causes the object to rotate about a fixed point, then a net torque is said to have been applied to the object. This net torque depends on the level arm, defined as the distance from the pivot point to the point the force is applied and the amount of that force applied at a right angle to the level arm (see Figure 1).

If the angle is anything other than 0°, then a net torque exists and rotation about the fixed point occurs. Counterclockwise rotation is designated as negative, while clockwise is labeled positive. If more than one force is acting on the lever arm, it is the sum of the torques that determines not only the extent, if any, of the rotation, but also its direction.

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 supporting bases (see Figure 2). 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.

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 3). The fulcrum of the truss is the pivot point where it is connected to the supporting wall.

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 (see Equations 2 and 3).

Experiment Overview

The lab begins with an introductory demonstration of the materials used in the guided-inquiry section of the investigation. The lever arm, the mass weight, the hangers, the spring scale and string are assembled on a support rod. The student is then instructed on taking data angles, lengths and forces. The experiment’s guided-inquiry design section presents a challenge to use the demonstration setup to model the task of hanging a “sign” over a sidewalk for the lowest material cost.

Materials

Prelab Questions

1. Use simple right-hand triangle geometry (see Figure 4) to determine the length of string (l) needed to support a 40-cm-long boom when the string is held at a 30° angle and the boom is held parallel with respect to the ground. Assume the string is attached to the wall directly above the pivot point of the boom.
   {13792_PreLab_Figure_4}
2. Using your solution to Question 1 as a guide, calculate the string lengths for the different string positions and string angles. Assume the string “attachment point” is directly above the pivot point of the boom. Record these values in a data table.

Safety Precautions

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

Procedure

Introductory Activity

1. Tie one end of a 50-cm string to the spring-scale handle.
   {13792_Procedure_Figure_5}
   {13792_Procedure_Figure_6}
2. The setup for the Simple Form Truss is shown in Figures 5 and 6.

   Note 1: The scale of the spring-scale should face the measurer.
   Note 2: 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.

   {13792_Procedure_Figure_7}

   Note 3: (Optional) Clamp the support stand to the tabletop using a C-clamp.
   Note 4: 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.

1. With the 500-g mass and the string attachment points fixed at given distances, does the force needed to maintain equilibrium decrease or increase when the string angle is increased? Explain.
2. If the distance from the pivot point is increased for the 500-g mass in Question 1 and the string is held at a fixed angle, does the force needed to maintain equilibrium increase? Explain.
3. Refer to Question 1. If the string attachment point is moved closer to the pivot point and the string angle is fixed, does the force needed to maintain equilibrium increase or decrease? Explain.
4. The purpose of this activity is to create a design for hanging a “sign” over a sidewalk for the lowest material cost. The parameters you are to work with are:
   • Sign mass: 500 g
   • Boom attachment height on wall: 20 cm
   • Maximum supporting weight of cable: 250 g (2.45 N)
   • Cost of supporting cable: $5/cm
5. Write a detailed step-by-step procedure to collect the data needed for optimizing your hanging sign model. Include all the materials and equipment that will be needed, along with all data tables. Once the data has been collected for a particular mass position and string angle, slowly lower the mass to the tabletop.

• Which boom setup completed the objectives in the best possible manner (i.e., sign hangs as far away from the wall as possible using the least amount of string)? How much did it cost for the cable?
• In order to provide the strongest support, where is the best place to hang the mass in relation to the string attachment?
• How does the angle of the string affect the tension in the string?>
• Compare the force measurements for when the string and mass are attached at the same position.
• Which boom in Figure 8 would produce the most tension in the string? Why? Assume the mass in each figure is the same.
  {13792_Procedure_Figure_8}

Student Worksheet PDF

13792_Student1.pdf