Elementary Algebra and Geometry for Schools

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HOME : Mathematics - Geometry

Introduction to lever 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?

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?

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?

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?

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.

Answers :

  • 1. 90lb.
  • 2. 4ft.
  • 3. 30lb.
  • 4. 7ft.
  • 5. 78lb.

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.

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?

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?

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?

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?

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.

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?

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?

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?

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.

Answers :

  • 1. Mike is 16 yr., sister is 12 yr.
  • 2. Mary weighs 90 lb.
  • 3. 17 nickels, 49 dimes.
  • 4. $4200 at 3%, $8400 at 5%.
  • 5. $7500 at 6%, $10,500 at 7%.
  • 6. Father is 36 yr., daughter is 9 yr.
  • 7. 18 nickels, 23 dimes, 9 quarters.
  • 8. 4½ ft.
  • 9. $3600 at 4½%, $4800 at 3%.
  • 10. Bob is 30 yr., John is 46 yr.

 

EXERCISES

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

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

3. Multiply -3a2b by 4ac.

4. Divide 45x33y6 by 15x2y2.

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

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

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

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

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

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

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

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

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

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

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?

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

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?

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.

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.

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?

Answers:

  • 1. 4x2
  • 2. 12a-6b.
  • 3. -12a33bc.
  • 4. 3y3.
  • 5. -17.
  • 6. -4.
  • 7. 2.
  • 8. 15x.
  • 9. 4x3 -5x2 -1.
  • 10. 2x -2y.
  • 11. x = (by - c)/a
  • 12. 7x + 2 = 5 + 4x.
  • 13. 600.
  • 14. 8.
  • 15. 3x + 3.
  • 16. 4x -3.
  • 17. .04x + .06y + .07z.
  • 18. 32, 160.
  • 19. 40 ft. width, 85 ft. length.
  • 20. 24 nickels, 27 quarters, 48 dimes.