Title: Pulleys and Belts

Problem: A man has 2 pulleys of radius 8in. and 3in. There is a belt that runs around both pulleys and the distance from the center of one pulley to the other is 13in. Find the total length of the belt. Draw a sketch of the problem, indicating the knowns and unknowns.

Math Topic/Concept: Using the Pythagorean theorem and trigonometric functions to solve a real world problem.

Materials: paper, pencil, compass, and calculator or trig. tables

Classroom Use: (Introductory/Developmental/Evaluation)Developmental and also Evaluation

Classroom use comments: This problem is a good test for using problem solving strategies and applying learned concepts. It also is an example of a real world problem using geometric and trigonometric concepts.

Grade: 11-12

Grade Cluster: Late HS

Illinois Goal: 9

Standard: A.5

Applied? (1-4): 4

Source: Dr. Jim Olsen

Answer: 62.5 inches long

Strategies Listed: To draw a parallel line, making a rectangle, to solve distances, and create a right triangle, then use the Pythagorean theorem and trigonometric functions to solve for more distances and angles

Solution: See Below

Other solution methods (if any):

Extensions or related problems: many possibilities

Intended rubric or assessment method: See below solution

Notes (if any): This problem brings in many concepts and puts them all together into one problem. It is a real test to see if the student can do this.

Write-up submitted by: Sandy Brooks

Solution:

Draw Line segment EB parallel to Line segment CD
CDBE is a rectangle
CA=8
CE=CB=3
AE=5
Use the Pythagorean theorem for rt. Triangle AEB
5^2 + EB^2 = 13^2
25 + EB^2 = 169
EB^2=144
EB = 12
EB=CD=XY=12
Find Arc CKX
First find the measure of angle EAB
COS angle EAB = 5/13 = 67.38…°
Angle CAX = 2 angle EAB = (67.38…) 2 = 134.76…°
Circumference of circle A = 2 pi r = 16 pi
Measure of Arc CKX = 360° - 134.76… = 225.2397°
Measure in inches of Arc CKX = 225.2397/360 x 16 pi = 31.449…in.
Similarly find Arc DLY
SIN angle EBA = 5/13 = 22.619…°
Angle ABD = 90° + 22.619° = 112.619…°
Arc DZY = 2 x 112.619 = 225.239…°
Arc DLY = 360 - 225.239 = 134.76°
C = 2 pi x 3 = 6 pi
Measure in inches of Arc DLY = 134.76/360 x 6 pi = 7.056… in.

Length of the Belt equals Arc CKX + Arc DLY + line CD + line XY
31.449 + 7.056 + 12 + 12 = 62.505
Belt is 62.5 inches long

Intended Assessment
Analytic Scoring Scale
For scoring student work on problem-solving problems
by James Olsen

Understanding the Problem 0 no attempt
1 attempt but little understanding
2 some understanding, some errors in understanding (1)
3 strong understanding
Planning & Solving 0 no attempt
2 inappropriate plan and not carried out
4 good plan, not carried out OR inappropriate plan and student carried out the plan
5 good plan, carried out, copying or calculation error (wrong answer)
6 good plan, carried out, correct answer
Extensions/checking/generalizations 0 no attempt (answer only)
1 Did a (one) look back and extend activity.
2 Did two or more look back and extend activities.
Jim Olson

Title:  Spinning Wheels

Problem:  The front wheels of a wagon measure 3.5 feet in diameter. The real wheels measure 4.25 feet in diameter. While the wagon is stopped somebody makes a chalk mark on both a front and rear wheel. How far must the wagon travel before both chalks marks return to their initial position at the same time?

Math Topic/Concept:  Geometry, Circumference of Circles, Least Common Multiples

Materials:  Paper, Pencil, Calculator optional

Classroom Use: (Developmental)

Classroom use comments: Review the formula for finding circumference of a circle.  Also could be used to introduce the concept of Least Common Multiples.

Grade:  7

Grade Cluster: (MS-Jr.High)

Illinois Goal:  9

Standard:  9.A.5

Applied? (1-4):  2

Source:  http://mathproblems.info/

Answer:  59.5*pi =~ 186.92 feet.

Strategies Listed:  Make a diagram, guess and check.

Solution:  The circumference of the front wheel is 3.5*pi and the circumference of the rear wheel is 4.25*pi. We must find the least common multiple of both these numbers. Both these numbers are divisible by .25*pi, so the least common multiple is .25*pi*LCM(14,17) = .25*pi*14*17 = 59.5*pi =~ 186.92.

Extensions or related problems:  The wheel diameters can be varied in order to get the students more accustomed to working with Least Common Multiples.

Intended rubric or assessment method:

1 point – using correct formula to find the circumference of the circle, given the diameter

1 point – finding the correct circumference

1 point – finding the least common multiple of the circumferences

1 point performing calculations correctly

Write-up submitted by:  Jason Carroll

James R. Olsen, Western Illinois University
E-mail: jr-olsen@wiu.edu
updated: June 30, 2005