UNDERSTANDING LEVERS by Mitch Ricketts Math Toolbox Is Designed to Help Readers Apply STEM Principles to Everyday Safety Issues

MATH TOOLBOX

UNDERSTANDING LEVERS By Mitch Ricketts Math Toolbox is designed to help readers apply STEM principles to everyday safety issues. Many readers may feel apprehensive about math and science. This series employs various communication strategies to make the learning process easier and more accessible.

This Math Toolbox article is the is a push or pull in a particular direction class levers are commonplace in anatomy, second of two installments on the role of (e.g., up, down, left, right, rotational). however, as illustrated by the human el- levers in occupational safety. In the first The magnitude (or amount) of force is bow, where the bicep exerts upward effort, installment (“The Case of the Tipping often specified in units of weight, such the elbow serves as the fulcrum, and the Forklift,” PSJ September 2020, pp. 45-51), as ounces, pounds or newtons. In a tendon attachment denotes the line of we considered the hazards associated first-class lever, the effort arm and load applied effort on the forearm. with out-of-control levers in workplaces. arm are located on opposite sides of the Recall from the first installment on This article will delve more deeply to un- fulcrum, and effort is applied in the same levers that the rotational force of a pivot- derstand the three main classes of levers, direction as the force exerted by the load. ing lever is known as torque (τ). Torque the concept of mechanical advantage and In a second-class lever (Figure 2), the is stated in units of distance times some associated calculations. effort arm and load arm are located on the weight, such as foot-pounds (ft-lb) and same side of the fulcrum. Effort is applied newton-meters (N-m). The direction of Classes of Levers in the opposite direction, compared with torque may be denoted with a positive Figure 1 depicts two examples of the force exerted by the load. Furthermore, sign for clockwise motion and a negative first-class levers: a stylized lever at top the load lies between the point of effort sign for counterclockwise rotation. As in and a pry bar as an everyday example at and the fulcrum. A wheelbarrow is an ev- the first installment, we will simplify by bottom. Remember from the first install- eryday example of a second-class lever. using the directionless, absolute value of ment that a lever is a rigid device that In a third-class lever (Figure 3), the ef- torque, designated as |τ|. pivots on a fulcrum. The fulcrum is the fort arm and load arm are located on the pivot point, or the point about which the same side of the fulcrum. The direction of Calculating Torque lever rotates. The lever has two “arms”: effort is opposite the direction of the force for Each Lever Class The load arm (or output arm) is the por- exerted by the load. Uniquely for third- In the first installment, we applied the tion of the lever directly connected to the class levers, the point of effort lies between formula for torque (absolute value) to load. The effort arm (or arm of applied the load and the fulcrum. Since these first-class levers as follows: force) is the portion of the lever to which levers tend to be inefficient, they are less we apply the effort, or input force. Force common in everyday technology. Third- |𝜏𝜏| = 𝑓𝑓 ∙ 𝑑𝑑 FIGURE 1 FIGURE 2 FIGURE 3 FIRST-CLASS SECOND-CLASS THIRD-CLASS LEVER EXAMPLES LEVER EXAMPLES LEVER EXAMPLES

52 PSJ PROFESSIONAL SAFETY OCTOBER 2020 assp.org FIGURE 4 OPPOSING TORQUES OF EFFORT & LOAD IN A FIRST-CLASS LEVER

FIGURE 5 EXAMPLE: BALANCED FIRST-CLASS LEVER

Example problem. Equivalent torques result in a balanced first-class lever. In this case, a 30-lb force applied to an effort arm of 1.5 ft balances the torque created by a 45-lb force applied to a load arm of 1 ft.

where: consider this 30-lb force to be applied at a •The weight applies a downward force |τ| = torque (absolute value); a turning point that represents the average location of 45 lb perpendicularly to the load arm. or twisting force where your hand presses on the lever, which This is the value of f2 in the formula. f = force applied perpendicularly (at a in this case is a distance of 1.5 ft to the left •The weight applies this force at an right angle) to the lever arm of the fulcrum. We will assume the weights average distance of 1 ft from the fulcrum. d = distance at which the force is ap- of the lever arms are negligible (or, alterna- This is the value of d2 in the formula. plied; the distance from the fulcrum to tively, that the weights of the lever arms are Based on these data, we calculate the line of applied force (measured per- balanced on both sides of the fulcrum). We torque generated by the load arm (|τload|) pendicularly to the line of applied force) will also assume friction is negligible. Under in ft-lb as follows: Note: When force is not applied per- these conditions, how many foot-pounds of Step 1: Start with the equation for pendicularly to the lever arm, a more torque are created by the effort arm to bal- torque of the load arm: general formula is used to account for the ance the torque created by the load arm? angle of applied force: τ = f ∙ d ∙ sin θ. We calculate torque created by the effort

Continuing our review, we can calculate arm (|τeffort|) based on the following data: #$%& ) ) torque for both sides of a first-class lever •Your hand applies a downward force Step 2: Insert|𝜏𝜏 the| known= 𝑓𝑓 ∙ 𝑑𝑑 values for as shown in Figure 4. For the effort arm, of 30 lb perpendicularly to the effort arm. weight of the load (f2 = 45 lb) and distance torque may be designated as |τeffort|. Force This is the value of f1 in the formula. (d2 = 1 ft). Then solve forτ | load| in ft-lb: and distance in the effort arm may be des- •Your hand applies this force at an av- ignated as f1 and d1, respectively. For the erage distance of 1.5 ft from the fulcrum. load arm, torque may be designated as |τload|, This is the value of d1 in the formula. |𝜏𝜏#$%&| = 𝑓𝑓) ∙ 𝑑𝑑) = 45 ∙ 1 = 45 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 with force and distance in the load arm Based on these data, we calculate Step 3: Our calculation indicates denoted as f2 and d2. If the opposing torques torque generated by the effort arm that the load arm is generating a torque are equal, then |τeffort| = |τload| and the lever (|τeffort|) in ft-lb as follows: of 45 ft-lb, which is equal to the torque will be balanced. On the other hand, if Step 1: Start with the equation for of the effort arm, meaning the torques torque created by the effort arm is greater torque of the effort arm: are balanced. In contrast, if the torques than torque created by the load arm, then had been unbalanced, the load would |τeffort| > |τload| and the load will rise. Finally, either rise (if |τeffort| > |τload|) or descend if torque created by the effort arm is less #$$%&' * * (if |τeffort| < |τload|). than torque created by the load arm, then Step 2: Insert!𝜏𝜏 the! known= 𝑓𝑓 ∙ 𝑑𝑑 values for Second-class lever calculated example: |τeffort| < |τload| and the load will descend. applied force (f1 = 30 lb) and distance (d1 We calculate torque for a second-class First-class lever calculated example: = 1.5 ft). Then solve forτ | effort| in ft-lb: lever using the same formula. However, Imagine you have used a first-class lever note that the effort and load arms over- to raise a load consisting of a 45-lb weight, lap because they are located on the same as illustrated in Figure 5. The load’s center #$$%&' * * side of the fulcrum, as shown in Figure of gravity is located 1 ft to the right of the !Step𝜏𝜏 3:! Our= 𝑓𝑓 ∙calculation𝑑𝑑 = 30 ∙ 1. 5indicates= 45 𝑓𝑓𝑓𝑓 −that𝑙𝑙𝑙𝑙 6 (p. 54). Also note that in second-class fulcrum. (Center of gravity is the average the effort arm is generating a torque of levers the effort is applied in a direction location of a body’s weight, or the location 45 ft-lb to balance the torque created by opposite to the direction of the force of where a body’s total weight is assumed to the load arm. the load. The value of d1 will be the entire be concentrated.) We will now calculate torque created distance from the fulcrum to the average To keep the load balanced (so it moves by the load arm (|τload|) to confirm that point of applied effort. The value of d2 neither up nor down), you exert a down- the torques are balanced, based on the will be the distance from the load’s center ward force of 30 lb with your hand. We will following data: of gravity to the fulcrum. Once again,

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the opposing torques are designated as force of 15 lb with your hand. Your Based on these data, we calculate |τeffort| and |τload|. Similarly, the effort and hand exerts this force at an average dis- torque generated by the effort arm load forces designated as f1 and f2, and tance of 2 ft to the left of the fulcrum. (|τeffort|) in ft-lb as follows: the distances are designated as d1 and d2 Assuming friction and the weights of Step 1: Start with the equation for for the effort and load arms respectively. the lever arms are negligible, how many torque of the effort arm: As always, |τeffort|=|τload| when the lever is foot-pounds of torque are created by the balanced; the load will rise when |τeffort| effort arm to balance the torque created > |τload| and the load will descend when by the load arm? !𝜏𝜏#$$%&'! = 𝑓𝑓* ∙ 𝑑𝑑* |τeffort| < |τload|. We calculate the torque created by the ef- Step 2: Insert the known values for To calculate opposing torques in a fort arm (|τeffort|) based on the following data: applied force (f1 = 15 lb) and distance (d1 second-class lever, imagine you have •Your hand applies an upward force of = 2 ft). Then solve forτ | effort| in ft-lb: used the lever shown in Figure 7 to raise 15 lb perpendicularly to the effort arm. a load consisting of a 40-lb weight. The This is the value of f1 in the formula. load’s center of gravity is located 0.75 ft •Your hand applies this force at an av- #$$%&' * * to the left of the fulcrum. To keep the erage distance of 2 ft from the fulcrum. Step!𝜏𝜏 3:! Our= 𝑓𝑓 calculation∙ 𝑑𝑑 = 15 ∙ 2 =indicates30 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 load balanced, you exert an upward This is the value of d1 in the formula. the effort arm is generating a torque

FIGURE 6 FIGURE 8 OPPOSING TORQUES OF EFFORT OPPOSING TORQUES OF EFFORT & LOAD IN A SECOND-CLASS LEVER & LOAD IN A THIRD-CLASS LEVER

FIGURE 7 FIGURE 9 EXAMPLE: BALANCED EXAMPLE: BALANCED SECOND-CLASS LEVER THIRD-CLASS LEVER

Example problem. Equivalent torques result in a balanced second-class Example problem. Equivalent torques result in a balanced third-class lever. In this case, a 15-lb force applied to an effort arm of 2 ft balances lever. In this case, a 37.5-lb force applied to an effort arm of 1.6 ft bal- the torque created by a 40-lb force applied to a load arm of 0.75 ft. ances the torque created by a 20-lb force applied to a load arm of 3 ft.

54 PSJ PROFESSIONAL SAFETY OCTOBER 2020 assp.org FIGURE 10 YOU DO THE MATH, PROBLEMS 1 & 4 of 30 ft-lb to balance the torque of the Based on these data, can calculate load arm. torque generated by the effort arm We will now calculate torque created by (|τeffort|) in ft-lb as follows: the load arm (|τload|) to confirm that the Step 1: Start with the equation for torques are balanced. Our data are as follows: torque of the effort arm: •The weight applies a downward force of 40 lb perpendicularly to the load arm. This is the value of f2 in the formula. !𝜏𝜏#$$%&'! = 𝑓𝑓* ∙ 𝑑𝑑* •The weight applies this force at an av- Step 2: Insert the known values for erage distance of 0.75 ft from the fulcrum. applied force (f1 = 37.5 lb) and distance FIGURE 11 This is the value of d2 in the formula. (d1 = 1.6 ft). Then solve forτ | effort| in ft-lb: Based on these data, we calculate YOU DO THE MATH, torque generated by the load arm (|τload|) PROBLEMS 2 & 5 in ft-lb as follows: #$$%&' * * Step 1: Start with the equation for !𝜏𝜏Step 3:! =Our𝑓𝑓 ∙calculation𝑑𝑑 = 37.5 ∙indicates1.6 = 60 that 𝑓𝑓𝑓𝑓 − the𝑙𝑙𝑙𝑙 torque of the load arm: effort arm is generating a torque of 60 ft-lb to balance the torque of the load arm. We will now calculate torque created by |𝜏𝜏#$%&| = 𝑓𝑓) ∙ 𝑑𝑑) the load arm (|τload|) to confirm that the Step 2: Insert the known values for torques are balanced. Our data are as follows: weight of the load (f2 = 40 lb) and distance •The weight applies a downward force (d2 = 0.75 ft). Then solve for |τload| in ft-lb: of 20 lb perpendicularly to the load arm. This is the value of f2 in the formula. •The weight applies this force at an average distance of 3 ft from the fulcrum. |𝜏𝜏#$%&| = 𝑓𝑓) ∙ 𝑑𝑑) = 40 ∙ 0.75 = 30 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 Step 3: Our calculation indicates that This is the value of d2 in the formula. FIGURE 12 the load arm is generating a torque of Based on these data, we calculate YOU DO THE MATH, 30 ft-lb, which is equal to the torque of the torque generated by the load arm (|τload|) effort arm, meaning the torques are bal- in ft-lb as follows: PROBLEMS 3 & 6 anced. In contrast, if the torques had been Step 1: Start with the equation for unbalanced, the load would either rise (if torque of the load arm: |τeffort| > |τload|) or descend (if |τeffort| < |τload|). Third-class lever calculated example: We use the same formula to calculate #$%& ) ) |𝜏𝜏 | = 𝑓𝑓 ∙ 𝑑𝑑 torque for a third-class lever. The effort Step 2: Insert the known values for and load arms overlap on the same side weight of the load (f2 = 20 lb) and distance of the fulcrum (as for second-class le- (d2 = 3 ft). Then solve for |τload| in ft-lb: vers), but in third-class levers the effort arm is always shorter than the load arm, as shown in Figure 8. |𝜏𝜏#$%&| = 𝑓𝑓) ∙ 𝑑𝑑) = 20 ∙ 3 = 60 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 To calculate opposing torques in a Step 3: Our calculation indicates tion that friction and the weights of the third-class lever, imagine you have used the load arm is generating a torque of lever arms are negligible: the lever shown in Figure 9 to raise a load 60 ft-lb, which is equal to the torque of a. How many foot-pounds of torque are consisting of a 20-lb weight. The load’s the effort arm. Note that the applied ef- created by the effort arm to balance the center of gravity is located 3 ft to the left fort (37.5 lb) is greater than the weight of torque created by the load arm? Use the of the fulcrum. To keep the load balanced, the load (20 lb) because the effort arm is equation for torque of the effort arm (|τeffort|) you exert an upward force of 37.5 lb with shorter than the load arm. and solve in units of foot-pounds (ft-lb). your hand. Your hand exerts this force at b. How many foot-pounds of torque are an average distance of 1.6 ft to the left of You Do the Math created by the load arm? Use the equation the fulcrum. Assuming friction and the Apply your knowledge to the following for torque of the load arm (|τload|) and weights of the lever arms are negligible, questions. Answers are on p. 63. solve in units of foot-pounds (ft-lb). how many foot-pounds of torque are 1. Imagine you have used a first-class 2. Imagine you have used a second-class created by the effort arm to balance the lever to raise a load consisting of a 100-lb lever to raise a load consisting of a 300-lb torque created by the load arm? weight, as illustrated in Figure 10. The weight, as illustrated in Figure 11. The We calculate torque created by the effort load’s center of gravity is located 2 ft to load’s center of gravity is located 0.5 ft arm (|τeffort|) based on the following data: the right of the fulcrum. To keep the to the left of the fulcrum. To keep the •Your hand applies an upward force of load balanced (so it moves neither up nor load balanced (so it moves neither up 37.5 lb perpendicularly to the effort arm. down), you exert a downward force of nor down), you exert an upward force of This is the value of f1 in the formula. 50 lb with your hand. Your hand exerts 60 lb with your hand. Your hand exerts •Your hand applies this force at an av- this force at an average distance of 4 ft to this force at an average distance of 2.5 ft to erage distance of 1.6 ft from the fulcrum. the left of the fulcrum. Answer the fol- the left of the fulcrum. Answer the follow- This is the value of d1 in the formula. lowing questions, based on the assump- ing questions, based on the assumption

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FIGURE 13 MECHANICAL ADVANTAGE

Mechanical advantage: Variables Fi, Fo, di and do for three classes of levers. For balanced levers (and when friction and the weights of the lever arms are negligible), the magnitudes of Fi and Fo are equal to the magnitudes of f1 and f2 from the earlier equations; however, the direction of Fo is opposite the direction of the previous f2. The values of di and do are equivalent to d1 and d2 from the previous equations. For unbalanced levers, the magnitude of Fo must either be measured with a scale or calculated as Fo = |τeffort| ÷ do or as Fo = MA ∙ Fi.

that friction and the weights of the lever values of mechanical advantage translate where: arms are negligible: to greater leverage. Specifically, the value MA = mechanical advantage, or amplifi- a. How many foot-pounds of torque are of MA will be greater than one (MA > 1) cation of force by a lever (ignoring friction, created by the effort arm to balance the when a machine’s output force is greater flexure and weight of lever components) torque created by the load arm? Use the than the input force. The value of MA Fi = input force; the force applied to the equation for torque of the effort arm (|τeffort|) will be equal to one (MA = 1) when a ma- effort arm; equivalent to f1 in the previous and solve in units of foot-pounds (ft-lb). chine’s output force equals the input force. calculations, assuming that friction and b. How many foot-pounds of torque are The value of MA will be less than one (MA the weights of the lever arms are negligible created by the load arm? Use the equation < 1) when a machine’s output force is less Fo = output force; the force output for torque of the load arm (|τload|) and than the input force. For example, if MA = through the load arm; for a lever in a solve in units of foot-pounds (ft-lb). 4, the machine’s output force is four times state of balance, and when friction and 3. Imagine you have used a third-class the input force; if MA = 2, the output force the weights of the lever arms are negli- lever to raise a load consisting of a 62-lb is two times the input force; if MA = 0.75, gible, Fo is equivalent in magnitude to weight, as illustrated in Figure 12 (p. 55). the output force is three-quarters the input f2 (but opposite in direction); for an un- The load’s center of gravity is located 3.1 ft force; if MA = 0.25, the output force is balanced lever, the magnitude of Fo must to the left of the fulcrum. To keep the load one-quarter the input force, and so forth. either be measured with a scale or calcu- balanced (so it moves neither up nor down), For levers, mechanical advantage can be lated as Fo = |τeffort| ÷ do or as Fo = MA ∙ Fi you exert an upward force of 155 lb with calculated in either of two ways: 1. as out- di = input distance; the distance from your hand. Your hand exerts this force at an put force (Fo) divided by input force (Fi); or the fulcrum to the point where input average distance of 1.24 ft to the left of the 2. as the length of the input arm (di) divided force is applied through the effort arm; fulcrum. Answer the following questions, by the length of the output arm (do). When equivalent to d1 based on the assumption that friction and the lever is balanced and when friction and do = output distance; the distance from the weights of the lever arms are negligible: the weights of the lever arms are negligible, the fulcrum to the point where output a. How many foot-pounds of torque are Fi and Fo are equivalent to f1 and f2 from force is applied through the load arm; created by the effort arm to balance the the previous equations. In other words, equivalent to d2 torque created by the load arm? Use the the input force (F1) is simply the effort we First-class lever calculated example: equation for torque of the effort arm (|τeffort|) apply to the lever and the output force (F2) Consider the first-class lever illustrated and solve in units of foot-pounds (ft-lb). has a magnitude equal to the weight of the in Figure 5 (p. 53). The lever is balanced. b. How many foot-pounds of torque are load (but a direction opposite that of the In other words, the lever is moving nei- created by the load arm? Use the equation weight of the load). Similarly, di and do are ther up nor down because the torques are for torque of the load arm (|τload|) and equivalent to d1 and d2 from the previous equivalent on both sides of the fulcrum. solve in units of foot-pounds (ft-lb). equations. The two equations are illustrated For balanced levers (assuming friction in Figure 13 and are defined as follows: and the weight of the lever arms are neg- Mechanical Advantage ligible), the magnitudes of Fi and Fo are Mechanical advantage (MA) is the am- equal to the magnitudes of f1 and f2, re- plification of force by a machine. Higher % & i o 𝐹𝐹 𝑑𝑑 spectively. Thus, F = 30 lb and F = 45 lb. 𝑀𝑀𝑀𝑀 = & 𝑜𝑜𝑜𝑜 𝑀𝑀𝑀𝑀 = % 56 PSJ PROFESSIONAL SAFETY OCTOBER 2020 assp.org𝐹𝐹 𝑑𝑑 Based on these data, we can calculate Next, we will calculate mechanical the mechanical advantage (MA) based on advantage of the lever in Figure 9 (p. 𝐹𝐹% 40 the forces Fi and Fo as follows: (rounded two places 54) based on the distances. Since di 𝑀𝑀𝑀𝑀 = & = = 2.67 Step 1: Start with the equation for me- 𝐹𝐹 15 beyond the decimal) and do are equivalent to d1 and d2, we chanical advantage based on force: see from Figure 9 that di = 1.6 ft and Step 3: Our calculation indicates that do = 3 ft. the lever has a mechanical advantage of Step 1: Start with the equation for 𝐹𝐹% 2.67. In other words, the lever amplifies mechanical advantage based on dis- 𝑀𝑀𝑀𝑀 = Step 2: Insert the known𝐹𝐹& values for the the input force by a multiple of 2.67. tance: input and output forces (Fi = 30 lb; Fo = Next, we will calculate mechanical 45 lb). Then solve for MA: advantage of the lever in Figure 7 (p. 54) based on the distances. Since di and do 𝑑𝑑% 𝑀𝑀𝑀𝑀 = are equivalent to d1 and d2, we see from 𝑑𝑑& % Step 2: Insert the known values for the 𝐹𝐹 45 Figure 7 that di = 2 ft and do = 0.75 ft. 𝑀𝑀𝑀𝑀 = = = 1.5 input and output distances (di = 1.6 ft; do 𝐹𝐹& 30 Step 1: Start with the equation for me- Step 3: Our calculation indicates that chanical advantage based on distance: = 3 ft). Then solve for MA: the lever has a mechanical advantage of 1.5. In other words, the lever amplifies the input force by a multiple of 1.5 (i.e., % 𝑑𝑑% 1.6 every 1 lb of force applied to the input, or 𝑑𝑑 𝑀𝑀𝑀𝑀 = = = 0(rounded.53 ) 𝑀𝑀𝑀𝑀 = & 𝑑𝑑& 3 effort, arm produces an output of 1.5 lb). Step 2: Insert the known𝑑𝑑 values for the Next, we will calculate mechanical input and output distances (di = 2 ft; do = Step 3: Our calculation indicates that advantage of the lever in Figure 5 (p. 53) 0.75 ft). Then solve for MA: the lever has a mechanical advantage based on the distances, di and do. Since of 0.53, confirming the result from the di and do are equivalent to d1 and d2 from equation based on force. the previous equations, di = 1.5 ft and do % 𝑑𝑑 2 (rounded) = 1 ft for the lever in Figure 5. 𝑀𝑀𝑀𝑀 = & = = 2.67 You Do the Math Step 1: Start with the equation for me- 𝑑𝑑 0.75 Apply your knowledge to the following chanical advantage based on distance: Step 3: Our calculation indicates that questions. Answers are on p. 63. the lever has a mechanical advantage of 4. Consider the first-class lever illus- 2.67, confirming the result we obtained trated in Figure 10 (p. 55). Since the lever % is balanced, the magnitudes of Fi and Fo 𝑑𝑑 using the equation based on force. 𝑀𝑀𝑀𝑀 = Third-class lever calculated ex- equal the magnitudes of f1 and f2, respec- 𝑑𝑑& Step 2: Insert the known values for the ample: Consider the third-class lever tively. Also, di and do are equivalent to d1 input and output distances (di = 1.5 ft; do shown in Figure 9 (p. 54). Since the and d2. Answer the following: = 1 ft). Then solve for MA: lever is balanced, the magnitudes of Fi a. Calculate the value of mechanical ad- and Fo equal the magnitudes of f1 and f2, vantage (MA) based on the forces Fi and Fo. b. Calculate the value of mechanical respectively. In other words, Fi = 37.5 lb and Fo = 20 lb. advantage (MA) based on the distances 𝑑𝑑% 1.5 𝑀𝑀𝑀𝑀 = & = = 1.5 Based on these data, we can calculate di and do. Step 3: Our calculation𝑑𝑑 1 indicates that the mechanical advantage (MA) based on 5. Consider the second-class lever the lever has a mechanical advantage of the forces as follows: illustrated in Figure 11 (p. 55). Since the 1.5, confirming the result we obtained Step 1: Start with the equation for me- lever is balanced, the magnitudes of Fi using the equation based on force. chanical advantage based on force: and Fo equal the magnitudes of f1 and f2, Second-class lever calculated example: respectively. Also, di and do are equiva- Consider the second-class lever shown lent to d1 and d2. Answer the following: in Figure 7 (p. 54). Since the lever is bal- 𝐹𝐹% a. Calculate the value of mechanical ad- anced, the magnitudes of Fi and Fo equal 𝑀𝑀𝑀𝑀 = vantage (MA) based on the forces Fi and Fo. 𝐹𝐹& the magnitudes of f1 and f2, respectively. In Step 2: Insert the known values for the b. Calculate the value of mechanical other words, Fi = 15 lb and Fo = 40 lb. input and output forces (Fi = 37.5 lb; Fo = advantage (MA) based on the distances Based on these data, we can calculate 20 lb). Then solve for MA: di and do. the mechanical advantage (MA) based on 6. Consider the third-class lever illus- the forces as follows: trated in Figure 12 (p. 55). Since the lever Step 1: Start with the equation for me- % is balanced, the magnitudes of Fi and Fo 𝐹𝐹 20 chanical advantage based on force: 𝑀𝑀𝑀𝑀 = = = 0(rounded.53 ) equal the magnitudes of f1 and f2, respec- 𝐹𝐹& 37.5 tively. Also, di and do are equivalent to d1 Step 3: Our calculation indicates that and d2. Answer the following: 𝐹𝐹% the lever has a mechanical advantage of a. Calculate the value of mechanical ad- 𝑀𝑀𝑀𝑀 = 𝐹𝐹& 0.53. In other words, the lever amplifies vantage (MA) based on the forces Fi and Fo. Step 2: Insert the known values for the the input force by a multiple of 0.53. Note b. Calculate the value of mechanical input and output forces (Fi = 15 lb; Fo = that since MA is less than one, the output advantage (MA) based on the distances 40 lb). Then solve for MA: force is less than the input force. di and do.

assp.org OCTOBER 2020 PROFESSIONAL SAFETY PSJ 57 MATH TOOLBOX

Summary & Limitations perpendicularly to the lever arms and the equation for torque of the effort arm (|τeffort|) In these two articles, we have explored levers were balanced. Additional vari- and solve in units of foot-pounds (ft-lb). basic concepts of leverage. We began in ables must be included in the equations b. How many foot-pounds of torque are the first installment by considering how when these assumptions are not met. created by the load arm? Use the equation out-of-control levers may cause serious for torque of the load arm (|τload|) and injuries in the workplace. We also calcu- How Much Have I Learned? solve in units of foot-pounds (ft-lb). lated torque as the product of force times Try these problems on your own. An- c. Calculate the value of mechanical distance. In this installment, we exam- swers are on p. 63. advantage (MA) based on the forces Fi ined the characteristics of three classes 7. Imagine you have used a first-class and Fo. Since the lever is balanced, the of levers and calculated mechanical lever to raise a load consisting of a 297-lb magnitudes of Fi and Fo equal the magni- advantage as the lever’s amplification of weight, as illustrated in Figure 14. The tudes of f1 and f2, respectively. force. Keep in mind that our calculations load’s center of gravity is located 1.3 ft d. Calculate the value of mechanical assumed friction and the weights of the to the right of the fulcrum. To keep the advantage (MA) based on the distances lever arms were negligible. Our calcula- load balanced (so it moves neither up di and do, which are equivalent to d1 and tions also assumed forces were applied nor down), you exert a downward force d2, respectively. of 110 lb with your hand. Your hand ex- 9. Imagine you have used a third-class FIGURE 14 erts this force at an average distance of lever to raise a load consisting of a 72-lb HOW MUCH HAVE I 3.51 ft to the left of the fulcrum. Answer weight, as illustrated in Figure 16. The LEARNED, PROBLEM 7 the following questions, based on the load’s center of gravity is located 2.75 ft assumption that friction and the weights to the left of the fulcrum. To keep the of the lever arms are negligible: load balanced (so it moves neither up a. How many foot-pounds of torque are nor down), you exert an upward force of created by the effort arm to balance the 120 lb with your hand. Your hand exerts torque created by the load arm? Use the this force at an average distance of 1.65 ft equation for torque of the effort arm (|τeffort|) to the left of the fulcrum. Answer the and solve in units of foot-pounds (ft-lb). following questions, based on the as- b. How many foot-pounds of torque are sumption that friction and the weights of created by the load arm? Use the equation the lever arms are negligible: for torque of the load arm (|τload|) and a. How many foot-pounds of torque are solve in units of foot-pounds (ft-lb). created by the effort arm to balance the c. Calculate the value of mechanical torque created by the load arm? Use the equation for torque of the effort arm (|τeffort|) FIGURE 15 advantage (MA) based on the forces Fi and Fo. Since the lever is balanced, the and solve in units of foot-pounds (ft-lb). HOW MUCH HAVE I magnitudes of Fi and Fo equal the magni- b. How many foot-pounds of torque are created by the load arm? Use the equation LEARNED, PROBLEM 8 tudes of f1 and f2, respectively. d. Calculate the value of mechanical for torque of the load arm (|τload|) and advantage (MA) based on the distances solve in units of foot-pounds (ft-lb). di and do, which are equivalent to d1 and c. Calculate the value of mechanical d2, respectively. advantage (MA) based on the forces Fi 8. Imagine you have used a second- and Fo. Since the lever is balanced, the class lever to raise a load consisting of magnitudes of Fi and Fo equal the magni- a 286-lb weight, as illustrated in Figure tudes of f1 and f2, respectively. 15. The load’s center of gravity is located d. Calculate the value of mechanical 1.4 ft to the left of the fulcrum. To keep advantage (MA) based on the distances the load balanced (so it moves neither di and do, which are equivalent to d1 and up nor down), you exert an upward force d2, respectively. of 130 lb with your hand. Your hand ex- For Further Study erts this force at an average distance of FIGURE 16 Learn more from the following source: 3.08 ft to the left of the fulcrum. Answer ASSP’s ASP Examination Prep: Program HOW MUCH HAVE I the following questions, based on the Review and Exam Preparation, edited by assumption that friction and the weights LEARNED, PROBLEM 9 Joel M. Haight, 2016. PSJ of the lever arms are negligible: a. How many foot-pounds of torque are References created by the effort arm to balance the Ricketts, M. (2020, Sept.). The case of the tip- torque created by the load arm? Use the ping forklift. Professional Safety, 65(9), 45-51.

Mitch Ricketts, Ph.D., CSP, is an associate professor of safety management at Northeastern State University (NSU) in Tahlequah, OK. He has worked in OSH since 1992, with experience in diverse settings such as agriculture, manufacturing, chemical/biological laboratories and school safety. Ricketts holds a Ph.D. in Cognitive and Human Factors Psychology from Kansas State University, an M.S. in Occupa- tional Safety Management from University of Central Missouri, and a B.S. in Education from Pittsburg State University. He is a professional member and officer of ASSP’s Tulsa Chapter, and faculty advisor for the Society’s NSU Broken Arrow Student Section.

58 PSJ PROFESSIONAL SAFETY OCTOBER 2020 assp.org board. The main switch was off, but a average number of missed workdays was References switch disconnects power on the load 2.3 times higher for truck driver injuries Associated Press. (2005, July 8). 450 Turkish side of the equipment, not the line side. (Gammon et al., 2019). sheep leap to their deaths. Fox News. www The incoming busses supplying power to For electrical work, driving and many .foxnews.com/story/450-turkish-sheep-leap the equipment from the utility were still common jobs, known hazards and hid- -to-their-deaths live. The electric company had not been den perils exist in work environments Amen, D.G. & Amen, T. (2016). The brain called to shut down utility power. The warrior’s way. Berkley. that seem safe. In sheeplike fashion, Divine, M. (with Machate, A.E.). (2013). The man, who was in his 20s, suffered a cat- workers are trained to recognize typical way of the SEAL: Think like an elite warrior to astrophic burn injury, and was disabled hazards in their respective workplaces. lead and succeed. Reader’s Digest. and disfigured. A description of what Yet, workers also need to be trained like Gammon, T., Lee, W.-J. & Intwari, I. (2019). the man physically had to endure made warriors to remain alert and vigilant Reframing our view of workplace “electrical” his deposition difficult to read, but I was about known and as-yet-unidentified injuries. IEEE Transactions on Industry Appli- most chilled by the emotional pain in dangers. Equally important, warriors cations, 55(4), 4370-4376. https://doi.org/10.11 his cry that no woman would ever want protect and aid others. In too many in- 09/TIA.2019.2907579 him. In the other incident, an apprentice cidents I have investigated, a worker was Grossman, D. (with Christensen, L.W.). died from severe burn injuries. The ap- injured in the presence of a coworker (2008). On combat: The psychology and physi- ology of deadly conflict in war and in peace (3rd prentice was working with a journeyman who was aware of the unsafe work prac- ed.). Warrior Science Publications. electrician on live equipment without tice but did not intervene. PSJ PPE when an arc flash occurred, which severely injured them both. Tammy Gammon, Ph.D., P.E., is a senior electrical engineer for John Matthews and Associates, a I tend to think of wolves disrupting consulting electrical engineering firm that specializes in electrical power systems, fires of electrical ori- a safe work environment for others. An gin, and electrical arc and shock injuries. She is the former research manager for the Arc-Flash Research example is a supervisor who does not Project, a collaborative effort between IEEE and National Fire Protection Association, and has authored many papers on electrical safety, electrical injuries and arc-flash hazards. She holds a bachelor’s and wear the required PPE. I have too often master’s degree and Ph.D. in electrical engineering from Georgia Institute of Technology. Gammon is a seen the behavior of lone wolves result professional member of ASSP’s Georgia Chapter. in needless electrical injury. One electrician was fatally shocked by high-voltage test equipment. Three glaring safety violations were committed: 1. He was Math Toolbox, continued from pp. 52-58 not wearing the appropriate gloves; How Much Have I Learned? 2. he was in contact with the equipment Answers: Understanding Levers under test; and 3. neither he (by choice) You Do the Math 7a. Your answers may vary slightly due to nor the helper were operating a safety !𝜏𝜏#$$%&'! = 𝑓𝑓* ∙ 𝑑𝑑* = 110 ∙ 3.51 = 386.1 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 rounding. switch. If any of these measures had 7b. been followed, the electrocution would 1a. #$%& ) ) |𝜏𝜏 | = 𝑓𝑓 ∙ 𝑑𝑑 = 297 ∙ 1.3 = 386.1 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 not have occurred. He and the team #$$%&' * * !𝜏𝜏 ! = 𝑓𝑓 ∙ 𝑑𝑑 = 50 ∙ 4 = 200 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 7c. leader had about 30 years of experience % 1b. 𝐹𝐹 297 𝑀𝑀𝑀𝑀 = = = 2.7 each and the helper had 10 years’ expe- 𝐹𝐹& 110 rience. In another incident, two elec- |𝜏𝜏#$%&| = 𝑓𝑓) ∙ 𝑑𝑑) = 100 ∙ 2 = 200 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 7d. 2a. 𝑑𝑑% 3.51 trical workers were investigating power 𝑀𝑀𝑀𝑀 = = = 2.7 𝑑𝑑& 1.3 loss to an industrial process and an arc !𝜏𝜏#$$%&'! = 𝑓𝑓* ∙ 𝑑𝑑* = 60 ∙ 2.5 = 150 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 8a. flash occurred. The severely burned 2b. #$$%&' * * worker was wearing a synthetic-fabric !𝜏𝜏 ! = 𝑓𝑓 ∙ 𝑑𝑑 = 130 ∙ 3.08 = 400.4 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 #$%& ) ) jacket with facility-issued PPE. |𝜏𝜏 | = 𝑓𝑓 ∙ 𝑑𝑑 = 300 ∙ 0.5 = 150 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 8b. 3a. Workplace injuries might be reduced if |𝜏𝜏#$%&| = 𝑓𝑓) ∙ 𝑑𝑑) = 286 ∙ 1.4 = 400.4 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 #$$%&' * * we could transcend the sheep-sheepdog- !𝜏𝜏 ! = 𝑓𝑓 ∙ 𝑑𝑑 = 155 ∙ 1.24 = 192.2 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 8c. wolf model and train workers to have 3b. 𝐹𝐹% 286 𝑀𝑀𝑀𝑀 = = = 2.2 the mindset of safety warriors. Some 𝐹𝐹& 130 |𝜏𝜏#$%&| = 𝑓𝑓) ∙ 𝑑𝑑) = 62 ∙ 3.1 = 192.2 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 8d. occupations are safer, but the risks of % 4a. 𝑑𝑑 3.08 𝑀𝑀𝑀𝑀 = = = 2.2 many common occupations are not fully 𝐹𝐹% 100 𝑑𝑑& 1.4 realized. For example, electrical work is 𝑀𝑀𝑀𝑀 = & = = 2 9a. 𝐹𝐹 50 4b. often perceived as a dangerous occupa- !𝜏𝜏#$$%&'! = 𝑓𝑓* ∙ 𝑑𝑑* = 120 ∙ 1.65 = 198 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 tion; electrical and nonelectrical workers 𝑑𝑑% 4 𝑀𝑀𝑀𝑀 = = = 2 9b. suffer electrical injuries. However, in 𝑑𝑑& 2 5a. #$%& ) ) |𝜏𝜏 | = 𝑓𝑓 ∙ 𝑑𝑑 = 72 ∙ 2.75 = 198 𝑓𝑓𝑓𝑓 − 𝑙𝑙𝑙𝑙 2016, the fatal injury rate for professional 𝐹𝐹% 300 𝑀𝑀𝐴𝐴 = & = = 5 9c. drivers was 24.9 (per 100,000 full-time 𝐹𝐹 60 % 5b. 𝐹𝐹 72 equivalent workers), while the fatal % 𝑀𝑀𝑀𝑀 = & = = 0.6 injury rate of electricians was 10. The 𝑑𝑑 2.5 𝐹𝐹 120 𝑀𝑀𝐴𝐴 = & = = 5 9d. 𝑑𝑑 0.5 % 2016 Bureau of Labor Statistics data also 6a. 𝑑𝑑 1.65 𝑀𝑀𝑀𝑀 = = = 0.6 show that the likelihood of injury was 𝐹𝐹% 62 𝑑𝑑& 2.75 𝑀𝑀𝑀𝑀 = = = 0.40 2.2 times higher for heavy-duty truckers 𝐹𝐹& 155 6b. (306) than for electricians (142) and the 𝑑𝑑% 1.24 𝑀𝑀𝑀𝑀 = & = = 0.40 𝑑𝑑 3.1assp.org OCTOBER 2020 PROFESSIONAL SAFETY PSJ 63