In this lesson, we will learn
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More Algebra Word Problems
More Algebra Topics
Lever Problems are word problems that use the lever principle. A good example of a lever is the seesaw.
A lever can be set up with unequal weights placed at different distances from the balancing point (also called the fulcrum).
In order for the above lever to be balanced, the following equation must be satisfied.
w1 × d1 = w2 × d2
Sketching a diagram of the situation would also be helpful in solving lever problems.
Example 1:
John weighs 90 lbs and Jane weighs 60 lbs. They are both seating on a seesaw. If John is seated 10 feet away from Jane, how far should each be from the fulcrum of the seesaw?
Solution:
Step 1: Assign variables:
Let x = distance of John from the fulcrum 10 – x = distance of Jane from the fulcrum
Step 2: Sketch the diagram
Step 3: Obtain equation from the diagram
90(x) = 60(10 – x)
Use the Distributive Property
90x = 600 – 60xIsolate variable x
90x + 60x = 600
150x = 600
x = 4
Answer: John should be 4 feet and Jane should be 6 feet from the fulcrum.
Example 2:
John wants to move a 400 lb. rock with a 5 ft. 9 in. crowbar. He puts the fulcrum 9 inches from the rock. How much force must he use to move the rock?
Solution:
Step 1: Assign variables :
Let x = force used
Change 5 ft. to 5 × 12 = 60 inches
Step 2: Sketch the diagram
Step 3: Obtain equation from the diagram
9 × 400 = 60 × x
Isolate variable x
60x = 3600
x = 60
Answer: He must use a force of 60 lbs.
Example 1:
John, Peter and Jane weigh 80, 60 and 50 lbs respectively. John sits 3 ft., Peter sits 5 ft. and Jane sits 6 ft. from the fulcrum on the same side. How far must their 200 lb. father sits from the fulcrum in order to balance them?
Solution:
Step 1: Assign variables :
Let x = distance of father from fulcrum
Step 2: Sketch the diagram
Step 3: Obtain equation from the diagram
200 × x = 3 × 80 + 5 × 60 + 6 × 50
Isolate variable x
200x = 240 + 300 + 300
200x = 840
x = 4.2
Answer: Their father must sit 4.2 ft. from the fulcrum.
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