Mathematics 5 Worksheet 2 Units and Proportion

Transcription

1 Mathematics 5 Worksheet 2 Units and Proportion Problem 1. Note that one yard is three feet. Assume that a day is exactly 24 hours. Water leaks from a faucet at a rate of.2 cubic yards per week. How many cubic inches does it leak per minute? Solution. The basic idea is that we can take any number or expression, and multiply it by 1, where we express 1 in a useful manner. For instance, we know that 1 week is the same as 7 days. That is, 1 week 7 days. Dividing both sides by 1 week, we obtain a conversion factor of 1 7days 1 week. Thus, writing 0.2 1/5, we have 1yard 3 5 week 1yard3 5 week week 7day day 24 hour hour 60 min ( ) 3 3 feet yard ( ) 3 12 in foot in 3 (after canceling units) min 3 3 (2 2 3) 3 in (2 3 3) (2 2 ( ) 3 5) min in (properties of exponents) 7 min 2 34 in (more properties of exponents) 7 min 162 in min. At ( ), we have factored each number into its prime factors (for instance, ). This is often a useful strategy for simplifying complicated fractions. Therefore the faucet leeks at a rate of 162/175 cubic inces per minute. If we don t mind approximating a bit, this is about 0.93 cubic inches per minute, or a bit less than 1 cubic inch per minute. 1

2 Problem 2. A painter needs two cubic feet of paint to paint a wall with an area of 2000 square feet. The painter now needs to paint a wall with an area of 2000 square yards. How many cubic yards of paint does he need? Solution. Note that if we double the area that we want to paint, then we need to double the volume of paint, and if we triple the area we want to paint, then we need triple the volume of paint. This implies that the ratio of paint needed (the volume) to the area covered is constant. That is volume constant. area Thus we may set up a proportion of the form 2ft ft 2 x yard yard 2, where x is the unknown volume of paint that we are trying to find. Solving for x, weget x yard 3 2ft yard ft 2ft yard 2 (simplify) ( 2ft yard ) yard 2 (multiply by 1) 3ft 2 3 yard3. Thus the painter would require 2/3 of a cubic yard of paint to cover 2000 square yards. 2

3 For the next two problems, remember that muscle strength and bone strength depend on the cross sectional area of the muscle and bone. Ignore complicating factors such as body mechanics. Problem 3. A giant stands 30 feet tall and has the exact same body proportions of a man who is five feet nine inches tall, weighs 170 pounds, and benchpresses 340 pounds. How much would you expect the giant to weigh? How much would you expect the giant to benchpress? What is the man s benchpress to bodyweight ratio? What is the giant s benchpress to body weight ratio? Solution. To compute weight, first consider a simpler example: the weight of a cube is equal to the product of its length, width, and height, times its density. So if a cube has side length s and density D, thenits weight (w o, for original weight ) is given by w o s 3 D. Suppose that we scale the size of the cube by a factor of C (for example, if C 2, we are doubling the size of the cube). Since we are scaling all of the lengths by C, but not changing the density, the weight of the new cube (w n for new weight ) is given by w n (Cs) 3 D. Thus we that the ratio of the old weight to the new weight is given by w n (Cs)3 D w o s 3 D C3 s 3 D s 3 D C3. Hence w n C 3 w o. In our problem, the scaling factor is the ratio of the giant s height to the man s height. That is, 30 ft C 5ft+9in in 5 12 in + 9 in Note that this scaling factor has no units. Then the giant s weight is given by ( ) w n C 3 w o 170 lbs. This expression, as ugly as it might be, is a perfectly reasonable answer (and is the answer that would be expected on an exam). However, it is not a very intuitive answer to get a better idea of what this number represents, we can use a calculator to get the approximation w n 24,000 lbs. 3

4 That is, the giant should weigh about 24,000 pounds. To determine how much the giant can benchpress, we use the reminder that strength is proportional to cross sectional area. For example, if we assume that a person s arm has a cross section that is a circle, then strength would be given by S o Kπr 2, where K is some constant that relates strength to area (is probably has some units that we don t know anything about), π is the ratio of a circle s diameter to its radius, and r is the radius of the person s arm. If we scale the person up by a factor of C, then we are scaling the radius up by a factor of C, and the strength of the scaled person would be S n Kπ(Cr) 2 C 2 (Kπr 2 )C 2 S o. As above, we know that C 120/ and we know that the man can benchpress 340 pounds, so we have S n ( ) lbs 9,300 lbs. Finally, we can compute the benchpress-to-bodyweight ratios. For the man, we have benchpress 340 lbs bodyweight 170 lbs 2, and for the giant, we have benchpress bodyweight ( 120 ( 120 ) lbs ) lbs In plain English, this says that we can expect the man to be able to benchpress about twice his own weight, while we can expect the giant to be able to benchpress just a little over one quarter of his own weight. 4

5 Problem 4. Suppose that a certain six foot tall man with a body weight of 180 pounds can squat 500 pounds. A mad scientist makes an exact copy of this man but scaled down to be one inch tall. How much does this copy weigh and how much can this copy squat? What is the squat weight to body weight ratio for the man and his miniature copy? Solution. The original person has a squat-to-weight ratio of 500 lbs 180 lbs As in problem 3, w n C 3 w o, where w o represents the weight of the original man, w n represents the weight of the shrunk man, and C is the ratio of their heights. In this case, w o is 180 pounds, and C 1in 6ft Therefore ( ) 3 1 w n 180 lbs lbs. 72 The copy can squat S n C 2 S o ( ) lbs lbs. 72 Hence the miniature person has a squat-to-weight ratio of ( S 1 ) 2 n lbs ( w n 1 ) lbs 180 Therefore the original man can be expected to squat a little less than three times his own weight, while the shrunk man should be able to squat 200 times his weight! 5