Have you ever been on a seesaw or opened a paint can's lid? If so, you were using a lever, which is a bar or other rigid object that has a point to pivot around called a fulcrum.
Let's say you are on a seesaw with your friend who weighs a lot more than you and you want the seesaw to be balanced. You could position yourself at a certain distance from the fulcrum so your lighter body weight counters your friend's body weight allowing the seesaw to be in equilibrium. This means there is no rotation. Let's see how the math works when dealing with a lever.
The cross product of force and distance is torque. The cross product is the mathematical process between two vectors that results in a vector perpendicular to both of the initial vectors. The law of the lever is also known as the law of moments and equates clockwise torques and counterclockwise torques. Equation 1 shows the law of levers.
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Imagine trying to open a paint can's lid with a screwdriver, but not applying the force perpendicularly to the screwdriver. If you applied the force at 1° to the screwdriver, it would take a lot of force to open the lid. Applying the force at 90° to the screwdriver would be the most effective at prying open the lid.
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This means we need a way to adjust the amount of torque based on the angle at which the force is applied to the lever. The sine trigonometric function has a maximum value at 90° and a minimum value at 0°. We can rewrite the cross product of force and distance from the lever arm including the sine function. This is how the cross product is evaluated. Equation 2 shows the equation for the magnitude of torques in the law of levers.
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We will deal with situations where the angle between the force and the lever arm is 90° so we can remove the sine functions because the sin of 90° is 1. This gives us Equation 3.
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Now, we can work some practice problems using the law of levers. Let's get to it!
Prompt: Imagine Sal weighs 150 pounds and his friend, Joe weighs 100 pounds. Sal sits 3 feet from the fulcrum on a seesaw. How far does Joe need to sit on the other side of the fulcrum so the seesaw is in equilibrium?
Solution: The first step in solving this problem is to draw a diagram, which we'll call Diagram 1.
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Now we can use Equation 3 to determine how far Joe needs to sit from the fulcrum so there is no rotation of the board they are sitting on.
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Now we can divide both sides by 100 lbs to get the distance Joe needs to sit from the fulcrum to put the seesaw in equilibrium. We'll call this Step 3.
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So if Joe sits anywhere besides 4.5 feet from the fulcrum, the board they are sitting on will rotate. Let's do another example.
Prompt: The lid on a paint can is wedged into the can. It would require 300 pounds of force to overcome the friction between the lid and the can. The screwdriver you have is 12 inches long. The fulcrum is 11 inches away from where you apply a downward force to pry open the lid, and the lid is 0.25 inches from the fulcrum. What minimum force would you need to apply to the screwdriver to open the can?
Solution: Again, we'll start the solution with a diagram.
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Now we can use Equation 3 to determine the force you need to apply perpendicularly to the lever arm to open the lid.
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Dividing both sides by 11 inches, we get the force you have to apply perpendicularly to the screwdriver.
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Roughly 6.8 pounds is not too bad compared to the 300 pounds you would have to apply to the lid to pull it off without the lever!
A lever is a rigid bar that has a point about which it can pivot called a fulcrum.
Force multiplied by the distance to the fulcrum, multiplied by the sine of the angle between the lever and the force is called torque. If the torques applied on opposite sides of the fulcrum are equal in magnitude and opposite in direction, the lever arm is at equilibrium, which means there will be no rotation about the fulcrum. In other words, the clockwise torques equal the counterclockwise torques.
The law of the lever equates torques on opposite sides of the fulcrum. If the angle between the force and lever arm is 90°, the sine function can be removed from the equation.
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High School Geometry: Homework Help Resource13 chapters | 142 lessons
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