Technical and student resources

Introduction to lever problems.

• See: Some Standard Types of Word Problems

A lever is a very simple "machine" which has many practical uses. It enables us to move heavy objects by using considerably lighter objects; it enables us to produce strong forces by exerting much smaller forces. In principle a lever works like the teeter or seesaw that you used to play on. You will remember that the teeter pivoted about a horizontal support.

A lever in simplest form is a bar that rests on a "knife-edge" support. In the adjacent figure the lever is represented by the line from A to B, which is supported on a "knife-edge," called the fulcrum, at F.

You will remember that if two children of equal weight sat at opposite ends of a teeter, they would balance each other; but if two children of unequal weight sat on the teeter, the heavier child had to move toward the fulcrum in order to balance the teeter. By means of accurate measurements it was discovered that a simple lever will be in balance if the product obtained by multiplying the weight on one side by the distance from the weight to the fulcrum is exactly equal to the product obtained by multiplying the weight on the other side by its distance to the fulcrum. In other words, if a child weighing 50 lb. sits 6 ft. from the fulcrum, he will balance a child weighing 75 lb. if this child sits on the other side of a seesaw at a distance of 4 ft. from the fulcrum, since (50)(6) is equal to (75)(4). This principle is stated by the formula:

w₁d₁ = w₂d₂

which is read, "w sub-one times d sub-one equals w sub-two times d sub-two." The formula indicates that a lever will be in balance if one weight (w₁) multiplied by its distance from the fulcrum (d₁) is equal to a second weight (w₂) multiplied by its distance from the fulcrum (d₂). In the notation w₁, the small 1 written to the right and below the letter is a subscript. Similarly the 1 in d, the 2 in w₂ and the 2 in d₂ are subscripts.

ILLUSTRATIVE EXAMPLE: lever problem

A weight of 120 lb. is placed 5 ft. from the fulcrum of a lever. Where should a weight of 50 lb. be placed to balance the lever?

Solution. We represent the number of feet from the fulcrum to the 50-lb. weight by x. Then we

Answering the question: The 50-lb. weight should be placed a distance of 12 ft. from the fulcrum and on the opposite side of the fulcrum from the 120-lb. weight.

EXERCISES: lever problems

1. A man weighing 150 lb. sits 6 ft. from the fulcrum of a seesaw and just balances his son on the other side who is seated 10 ft. from the fulcrum. How many pounds does the son weigh?

We can use the principle of moments to solve this problem. The principle of moments states that the sum of the moments acting on a system must be zero for the system to be in equilibrium. In this case, the seesaw is in equilibrium, so the sum of the moments acting on it must be zero.

The moment of a force is the product of the force and the perpendicular distance from the force to the fulcrum. In this case, the forces acting on the seesaw are the weight of the man and the weight of the son.

Let's assume that the fulcrum is at position 0 on a number line. The man is sitting at position 6, and the son is sitting at position -10 (since he is on the other side of the fulcrum). The weight of the man (150 lb) is acting downwards at position 6, and the weight of the son (unknown) is acting downwards at position -10.

The sum of the moments acting on the seesaw is:

150 lb * 6 ft + weight of son * (-10 ft) = 0

Simplifying and solving for the weight of the son:

150 lb * 6 ft = weight of son * 10 ft

Weight of son = 150 lb * 6 ft / 10 ft

Weight of son = 90 lb

Therefore, the son weighs 90 pounds.

2. Betty, who weighs 64 lb., sits 5 ft. from the fulcrum of a teeter. If Ralph weighs 80 lb., how far from the fulcrum must he sit in order to balance the teeter?

To balance the teeter, the total moment on each side of the fulcrum must be equal. The moment is the weight multiplied by the distance from the fulcrum. Let x be the distance Ralph sits from the fulcrum, in feet. Then we can write:

moment on Betty's side = moment on Ralph's side

64 lb * 5 ft = 80 lb * x ft

Simplifying:

320 = 80x

x = 320/80

x = 4

Therefore, Ralph must sit 4 feet from the fulcrum to balance the teeter, assuming the teeter is uniform and symmetrical.

3. A weight of 32 lb. is placed 7½ ft. from the fulcrum of a lever. To blance the lever, what weight should be placed on the other side at a distance of 8 ft. from the fulcrum?

To balance the lever, the total moment on each side of the fulcrum must be equal. The moment is the weight multiplied by the distance from the fulcrum. Let w be the weight that should be placed on the other side to balance the lever, in pounds. Then we can write:

moment on the left side = moment on the right side

32 lb * 7.5 ft = w * 8 ft

Simplifying:

240 = 8w

w = 240/8

w = 30

Therefore, a weight of 30 pounds should be placed 8 feet from the fulcrum to balance the lever.

4. A lever 10 ft. long is to be balanced by a weight of 36 lb. on one end a weight of 84 lb. on the other end. How far from the 36 lb. weight should the fulcrum be placed?

To balance the lever, the total moment on each side of the fulcrum must be equal. The moment is the weight multiplied by the distance from the fulcrum. Let x be the distance from the fulcrum to the 36 lb weight, in feet. Then we can write:

moment on the left side = moment on the right side

36 lb * x ft = 84 lb * (10 ft - x)

Simplifying:

36x = 840 - 84x

120x = 840

x = 7

Therefore, the fulcrum should be placed 7 feet from the 36 lb weight to balance the lever.

5. The sum of two weights is 195 lb. One weight is placed 36 in. from the fulcrum of a lever, and the other is placed on the other side of the fulcrum of the lever 24 in. from the fulcrum. If the lever is in balance, find the number of pounds in the lighter weight.

To balance the lever, the total moment on each side of the fulcrum must be equal. The moment is the weight multiplied by the distance from the fulcrum. Let x be the weight of the lighter weight, in pounds. Then we can write:

moment on the left side = moment on the right side

x lb * 36 in = (195-x) lb * 24 in

Simplifying:

36x = 4680 - 24x

60x = 4680

x = 78

Therefore, the weight of the lighter weight is 78 pounds.

CHAPTER REVIEW EXERCISES

1. Mike is 4 years older than his sister. Eight years ago, he was twice as old as she was then. Find their present ages.

Let's use algebra to solve the problem. We can start by defining some variables:

• Let's call Mike's current age "M"
• Let's call his sister's current age "S"

From the problem statement, we know that:

• M = S + 4 (Mike is 4 years older than his sister)
• 2(S - 8) = M - 8 (Eight years ago, Mike was twice as old as his sister was then)

We can use substitution to eliminate M from the second equation:

• 2(S - 8) = (S + 4) - 8
• 2S - 16 = S - 4
• S = 20

So Mike's sister is currently 20 years old. We can use the first equation to find Mike's age:

• M = S + 4
• M = 20 + 4
• M = 24

So Mike is currently 24 years old.

2. Mary weighs 30 lb. more than her younger sister. If Mary sits 4 ft. from the fulcrum of a seesaw, she just balances her sister when the sister sits 6 ft. from the fulcrum. How much does Mary weigh?

Let's call the weight of Mary's younger sister "S" (in pounds). Then we know that Mary's weight is S + 30.

To balance the seesaw, the total moment on each side of the fulcrum must be equal. The moment is the weight multiplied by the distance from the fulcrum. If we let x be the weight of Mary (in pounds), we can write:

moment on Mary's side = moment on sister's side

x * 4 ft = (S + 30) * 6 ft

Simplifying:

4x = 6S + 180

We also know that Mary's weight is S + 30, so we can substitute this expression for x:

4(S + 30) = 6S + 180

Simplifying:

4S + 120 = 6S + 180

2S = 60

S = 30

So Mary's younger sister weighs 30 pounds. Mary's weight is S + 30, or 60 pounds.

3. Steve has $5.75 in nickels and dimes. If the number of dimes is 2 less than three times the number of nickels, how many coins of each kind has he?

Let's use algebra to solve the problem. We can start by defining some variables:

• Let's call the number of nickels "N"
• Let's call the number of dimes "D"

From the problem statement, we know that:

• D = 3N - 2 (the number of dimes is 2 less than three times the number of nickels)
• 0.05N + 0.10D = 5.75 (Steve has $5.75 in nickels and dimes)

We can use substitution to eliminate D from the second equation:

• 0.05N + 0.10(3N - 2) = 5.75
• 0.05N + 0.30N - 0.20 = 5.75
• 0.35N = 5.95
• N = 17

So Steve has 17 nickels. We can use the first equation to find the number of dimes:

• D = 3N - 2
• D = 3(17) - 2
• D = 49

So Steve has 49 dimes.

4. Jeff invests a certain amount of money at 3% and twice the amount at 5%. If his annual income from these two investments is $546, how much has he invested at each rate?

Let's use algebra to solve the problem. We can start by defining some variables:

• Let's call the amount Jeff invested at 3% "x"
• Then the amount he invested at 5% is "2x" (twice the amount invested at 3%)

From the problem statement, we know that:

• The annual income from the 3% investment is 0.03x
• The annual income from the 5% investment is 0.05(2x) = 0.10x
• The total annual income is $546, so we can write:

0.03x + 0.10x = 546

Simplifying:

0.13x = 546

x = 4200

So Jeff invested $4,200 at 3% and $8,400 (twice as much) at 5%.

5. Mr. Miller invests $18,000, part at 6% and the rest at 7%. If his annual income from these two investments is $1185, how much has he invested at each rate?

Let's use algebra to solve the problem. We can start by defining some variables:

• Let's call the amount Mr. Miller invested at 6% "x"
• Then the amount he invested at 7% is "18000 - x" (the rest of the $18,000)

From the problem statement, we know that:

• The annual income from the 6% investment is 0.06x
• The annual income from the 7% investment is 0.07(18000 - x) = 1260 - 0.07x
• The total annual income is $1185, so we can write:

0.06x + 1260 - 0.07x = 1185

Simplifying:

-0.01x = -75

x = 7500

So Mr. Miller invested $7,500 at 6% and $10,500 ($18,000 - $7,500) at 7%.

6. A father is four times as old as his daughter. In 15 years the sum of their ages will be 75 years. Find he present age of each.

Let's use algebra to solve the problem. We can start by defining some variables:

• Let's call the daughter's age "x"
• Then the father's age is "4x" (four times the daughter's age)

From the problem statement, we know that:

• In 15 years, the daughter's age will be "x + 15"
• In 15 years, the father's age will be "4x + 15"
• The sum of their ages in 15 years will be 75, so we can write:

(x + 15) + (4x + 15) = 75

Simplifying:

5x + 30 = 75

5x = 45

x = 9

So the daughter is currently 9 years old. The father is four times as old, or 36 years old.

7. Jean has $.5.45 consisting of nickels, dimes, and quarters. She has twice as many nickels as quarters, and 5 more dimes than nickels. How many coins of each kind has she?

Let's use algebra to solve the problem. We can start by defining some variables:

• Let's call the number of quarters Jean has "q"
• Then the number of nickels she has is "2q" (twice as many as quarters)
• And the number of dimes she has is "2q + 5" (5 more than nickels)

From the problem statement, we know that:

• The value of one quarter is $0.25, so the value of "q" quarters is 0.25q dollars
• The value of one dime is $0.10, so the value of "2q + 5" dimes is 0.10(2q + 5) dollars
• The value of one nickel is $0.05, so the value of "2q" nickels is 0.05(2q) dollars
• The total value of all coins is $5.45, so we can write:

0.25q + 0.10(2q + 5) + 0.05(2q) = 5.45

Simplifying:

0.25q + 0.20q + 0.50 + 0.10q = 5.45

0.55q + 0.50 = 5.45

0.55q = 4.95

q = 9

So Jean has 9 quarters, 18 nickels (2q) and 23 dimes (2q + 5).

8. A lever 12 ft. long is to be balanced by a weight 100 lb. on one end and a weight of 60 lb. on the other end. How far from the 100 pound weight should the fulcrum be placed?

Let's call the distance from the fulcrum to the 100 lb weight "x", and the distance from the fulcrum to the 60 lb weight "12-x". To balance the lever, the product of the weight and its distance from the fulcrum must be the same on both sides. So we can write:

100x = 60(12-x)

Expanding and simplifying:

100x = 720 - 60x

160x = 720

x = 4.5

So the fulcrum should be placed 4.5 ft from the 100 lb weight.

9. A man invests a sum of money at 4½%, and a second sum $1200 greater than the first at 3%. If the annual income from the 4½% investment is $18 larger than the annual income from the 3% investment, how much has he invested at each rate?

Let's call the amount invested at 4.5% "x". Then the amount invested at 3% is "x + $1200".

The annual income from the 4.5% investment is equal to 4.5% of the amount invested, or 0.045x. The annual income from the 3% investment is equal to 3% of the amount invested, or 0.03(x + $1200).

We're told that the annual income from the 4.5% investment is $18 larger than the annual income from the 3% investment, so we can set up an equation:

0.045x = 0.03(x + $1200) + $18

Simplifying and solving for x:

0.045x = 0.03x + $36 + $18 0.015x = $54 x = $3600

So the man invested $3600 at 4.5% and $4800 ($3600 + $1200) at 3%.

10. Twelve years ago John was 2 years less than twice as old as Bob was then. Ten years from now the sum of their ages will be 96 years. Find their present ages.

Let's start by using algebra to translate the given information into equations we can solve.

Let's call John's current age "J" and Bob's current age "B". Then we can set up two equations based on the information given:

Twelve years ago John was 2 years less than twice as old as Bob was then:

J - 12 = 2(B - 12) - 2

Ten years from now the sum of their ages will be 96 years:

(J + 10) + (B + 10) = 96

Now we can solve for J and B by simplifying and combining like terms:

J - 12 = 2B - 26 J = 2B - 14

(J + 10) + (B + 10) = 96 J + B + 20 = 96 J + B = 76

Substituting J = 2B - 14 from the first equation into the second equation, we get:

(2B - 14) + B = 76 3B = 90 B = 30

So Bob is currently 30 years old. Substituting this value back into J = 2B - 14, we get:

J = 2(30) - 14 J = 46

So John is currently 46 years old.

EXERCISES

1. Find the sum of 7x², 9x², and -12x²

7x² + 9x² - 12x² can be simplified by combining the coefficients of x²:

(7 + 9 - 12)x² = 4x²

Therefore, the sum of 7x², 9x², and -12x² is 4x².

2. By how much does 15a -2b exceed 3a +4b?

To find out how much 15a - 2b exceeds 3a + 4b, we need to subtract the second expression from the first:

(15a - 2b) - (3a + 4b)

Simplifying the expression by removing the parentheses and combining like terms, we get:

15a - 2b - 3a - 4b = 12a - 6b

Therefore, 15a - 2b exceeds 3a + 4b by 12a - 6b.

3. Multiply -3a²b by 4ac.

-3a²b multiplied by 4ac is:

-3a²b * 4ac = (-3 * 4 * a² * a * c * b)

Simplifying:

-3a²b * 4ac = -12a³bc

4. Divide 45x³3y⁶ by 15x²y².

To divide (45x³)(3y⁶) by (15x²)(y²), we can simplify the expression by dividing the coefficients and the variables separately:

Coefficient Division:

(45*3) / (15) = 135/15 = 9

Variable Division:

x³ / x² = x^(3-2) = x

y⁶ / y² = y^(6-2) = y⁴

Putting it together, we get:

(45x³)(3y⁶) / (15x²)(y²) = 9x*y⁴

Simplifying further, we can write:

9xy⁴

Therefore, the quotient of (45x³)(3y⁶) divided by (15x²)(y²) is 9xy⁴.

5. If a =-2 and b = 3, find the value of a² -4ab -5b²

Substituting a=-2 and b=3 into the expression a² - 4ab - 5b², we get:

a² - 4ab - 5b² = (-2)² - 4(-2)(3) - 5(3)² = 4 + 24 - 45 = -17

Therefore, the value of a² - 4ab - 5b² when a=-2 and b=3 is -17.

6. Solve for y: 4-3y = 16.

Starting with the equation 4 - 3y = 16:

First, we isolate the variable term on one side of the equation by subtracting 4 from both sides:

4 - 3y - 4 = 16 - 4

Simplifying the left side by combining like terms:

-3y = 12

Finally, we isolate y by dividing both sides by -3:

y = -4

Therefore, the solution to the equation 4-3y=16 is y = -4.

7. Solve for x: 3(2 -x) +6 = 2(x + 1).

Starting with the equation 3(2 - x) + 6 = 2(x + 1):

First, we simplify both sides of the equation by distributing the coefficients:

6 - 3x + 6 = 2x + 2

Next, we combine like terms on each side of the equation:

12 - 3x = 2x + 2

We isolate the variable term on one side of the equation by adding 3x to both sides:

12 = 5x + 2

Then we isolate the variable by subtracting 2 from both sides:

10 = 5x

Finally, we isolate x by dividing both sides by 5:

x = 2

Therefore, the solution to the equation 3(2 - x) + 6 = 2(x + 1) is x = 2.

8. If you have 3x nickels, how would you represent the value of your nickels in cents?

Each nickel has a value of 5 cents.

Therefore, the total value of 3x nickels in cents is:

3x * 5 cents/nickel = 15x cents

So if you have 3x nickels, the value in cents would be 15x.

9. Subtract 5x² -x³ +2x -4 from 3x³ +2x -5.

(3x³ + 2x - 5) - (5x² - x³ + 2x - 4) = 4x³ - 5x² - 1

Here is the correct solution:

(3x³ + 2x - 5) - (5x² - x³ + 2x - 4)

Multiplying the second set of parentheses by -1 to distribute the negative sign:

3x³ + 2x - 5 - 5x² + x³ - 2x + 4

Rearranging the terms in descending order of degree:

3x³ + x³ - 5x² + 2x - 2x - 5 + 4

Combining like terms:

4x³ - 5x² - 1

Therefore, the result of subtracting (5x² - x³ + 2x - 4) from (3x³ + 2x - 5) is 4x³ - 5x² - 1.

10. Simplify: x -[y -(x -y)].

Simplifying x - [y - (x - y)] involves applying the distributive property and combining like terms:

x - [y - (x - y)] = x - [y - x + y] (distributing the negative sign) = x - y + x - y (combining like terms) = 2x - 2y

Therefore, x - [y - (x - y)] simplifies to 2x - 2y.

11. Solve for x in terms of the other letters: ax -by +c = 0.

To solve for x in terms of the other letters, we need to isolate x on one side of the equation:

ax - by + c = 0

Subtracting c from both sides:

ax - by = -c

Adding by to both sides:

ax = by - c

Dividing both sides by a:

x = (by - c)/a

Therefore, x in terms of the other letters is x = (by - c)/a.

12. Form an equation from the following statement, "7x +2 is 5 larger than 4x."

The given statement can be written as an equation as follows:

7x + 2 = 4x + 5

To see why this is the case, notice that the statement says "7x + 2 is 5 larger than 4x".

We can translate "7x + 2 is 5 larger than" to "7x + 2 =" plus "5".

Then we can translate "4x" to the expression on the right side of the equation.

Therefore, we end up with the equation 7x + 2 = 4x + 5.

13. Solve the equation: .05x + .06(2000 -x) = 114.

To solve the equation:

0.05x + 0.06(2000 - x) = 114

We can simplify and solve for x:

0.05x + 120 - 0.06x = 114 (distributing 0.06) -0.01x + 120 = 114 (combining like terms)

Subtracting 120 from both sides:

-0.01x = -6

Dividing both sides by -0.01:

x = 600

Therefore, the solution to the equation is x = 600.

14. Solve the equation: 3(5x -8)-2(4x -9) = 50.

To solve the equation:

3(5x - 8) - 2(4x - 9) = 50

We can simplify and solve for x:

15x - 24 - 8x + 18 = 50 (distributing the 3 and -2) 7x - 6 = 50 (combining like terms)

Adding 6 to both sides:

7x = 56

Dividing both sides by 7:

x = 8

Therefore, the solution to the equation is x = 8.

15. The number of years in a boy's age is represented by 3x -12. How would you represent the boy's age in 15 years?

If the number of years in a boy's age is represented by 3x - 12, then the boy's age now can be expressed as 3x - 12. To represent the boy's age in 15 years, we need to add 15 to the current age:

3x - 12 + 15

Simplifying:

3x + 3

Therefore, the boy's age in 15 years can be represented as 3x + 3.

16. If 4x -5 represents an even integer, how would you represent the next larger consecutive even integer?

If 4x - 5 represents an even integer, then we know that 4x - 5 is divisible by 2. In other words, 4x - 5 must be of the form 2n, where n is an integer.

To find the next consecutive even integer, we add 2 to the current even integer (which is equivalent to adding 1 to n in 2n).

So the next even integer can be represented as:

2 + 4x - 5

Simplifying:

4x - 3

Therefore, the next larger consecutive even integer can be represented as 4x - 3.

17. A man invests x dollars at 4%, y dollars at 6%, and z dollars at 7%. How would you represent his total annual income from these three investments?

The amount of annual income earned from an investment is equal to the product of the amount invested, the interest rate, and the duration of the investment (usually measured in years).

Therefore, the annual income earned from the investment of x dollars at 4% is:

0.04x

Similarly, the annual income earned from the investment of y dollars at 6% is:

0.06y

And the annual income earned from the investment of z dollars at 7% is:

0.07z

To find the total annual income earned from all three investments, we can add up these individual incomes:

Total annual income = 0.04x + 0.06y + 0.07z

18. One number is five times another number. Twice the smaller number, increased by 30, is equal to the larger number, decreased by 66. Find the two numbers.

Let's use algebra to solve the problem.

Let's call the smaller number "x" and the larger number "y".

From the problem statement, we know that:

y = 5x (one number is five times another number)

2x + 30 = y - 66 (twice the smaller number, increased by 30, is equal to the larger number, decreased by 66)

We can substitute the first equation into the second equation to eliminate "y" and solve for "x":

2x + 30 = 5x - 66

Adding 66 to both sides:

2x + 96 = 5x

Subtracting 2x from both sides:

96 = 3x

Dividing both sides by 3:

x = 32

Now that we know x = 32, we can use the first equation to find y:

y = 5x

y = 5(32)

y = 160

Therefore, the two numbers are 32 and 160.

19. The length of a rectangle is 5 ft. more than twice the width. If the perimeter of the rectangle is 250 ft., find its length and width.

Let's use algebra to solve the problem.

Let's call the width of the rectangle "w". Then, according to the problem statement, the length of the rectangle is "5 ft. more than twice the width", which we can express as:

length = 2w + 5

The perimeter of a rectangle is the sum of the lengths of all its sides, which for a rectangle is twice the length plus twice the width. So we can set up an equation for the perimeter of the rectangle:

perimeter = 2(length + width) = 2(2w + 5 + w) = 2(3w + 5) = 6w + 10

We are told that the perimeter is 250 ft., so we can set this equal to the expression we just derived:

6w + 10 = 250

Subtracting 10 from both sides:

6w = 240

Dividing both sides by 6:

w = 40

Now that we know the width is 40 ft., we can use the expression for the length we derived earlier to find the length:

length = 2w + 5 = 2(40) + 5 = 85

Therefore, the length and width of the rectangle are 85 ft. and 40 ft., respectively.

20. A man has 3 more quarters than nickels, and twice as many dimes as nickels. If the value of these coins is $12.75, how many coins of each kind has he?

Let's use algebra to solve the problem.

Let's call the number of nickels the man has "n". Then, according to the problem statement:

• The man has "3 more quarters than nickels", which means he has n + 3 quarters.
• The man has "twice as many dimes as nickels", which means he has 2n dimes.

We can now set up an equation for the value of the coins the man has:

0.05n + 0.10(2n) + 0.25(n + 3) = 12.75

Simplifying and solving for "n":

0.05n + 0.20n + 0.25n + 0.75 = 12.75

0.50n = 12.00

n = 24

Now that we know the man has 24 nickels, we can use the expressions we derived earlier to find the number of quarters and dimes:

• The man has n + 3 quarters, so he has 24 + 3 = 27 quarters.
• The man has 2n dimes, so he has 2(24) = 48 dimes.

Therefore, the man has 24 nickels, 27 quarters, and 48 dimes.