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Get Full Access to Engineering Mechanics: Statics & Dynamics - 14 Edition - Chapter 13 - Problem 13-44
Get Full Access to Engineering Mechanics: Statics & Dynamics - 14 Edition - Chapter 13 - Problem 13-44

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# If the motor draws in the cable with an acceleration of 3 ISBN: 9780133951929 123

## Solution for problem 13-44 Chapter 13

Engineering Mechanics: Statics & Dynamics | 14th Edition

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Problem 13-44

If the motor draws in the cable with an acceleration of 3 m>s 2 , determine the reactions at the supports A and B. The beam has a uniform mass of 30 kg>m, and the crate has a mass of 200 kg. Neglect the mass of the motor and pulleys.

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Koshar Amy Brogan March 6 & 8, 2016 Week 13 Cryptography Part 2 Review  Shift Cipher o A code that replaces letters with other letters Code the message “Send Ammo” using the Vignere key “SHOE” (18-7-14-4) S E N D A M M O 18 4 13 3 0 12 12 14 S H O E S H O E 18 7 14 4 18 7 14 4 36 11 27 7 18 19 26 18 10 1 0 K L B H S T A S Now that we have the coded message, try to decode it like as if you received it and needed to read it without knowing what it said. K L B H S T A S 10 11 1 7 18 19 0 18 At this point to work backwards, we need to subtract the code word from what we have, but some of the numbers won’t be big enough sowe have to add 26. K L B H S T A S 10 11 1 7 18 19 0 18 +26 = 36 +26 = 27 +26 = 26 S H O E S H O E -18 -7 -14 -4 -18 -7 -14 -4 18 4 13 3 0 12 12 14 S E N D A M M O Decode this: Key: OUT T I K K U K R Modular Arithmetic  Addition operation o Sum after adding  Multiplication operation o Product after repeated adding  Subtraction operation o Difference after subtracting  Division operation o Quotient after repeated subtracting  Modular Arithmetic o Remainder after dividing With Modular Arithmetic, the answer will be what is left over after dividing the value in question. It doesn’t matter how many timesa number goes into another, only what is left over. With small numbers this is easiest to see using long division Ex 1: 1 = 3 mod2  2 goes into 3 and there is 1 left over (3-2=1) Ex 2: 23 mod5 = 3 20 / 5, 3 left over Ex 3: mod 5 = 0, 1, 2, 3, 4 (all numbers before the value is 5) Ex 4: 27 mod4 = 3 27/4, 3 left over Ex 5: 66 mod11 =0  66 = 11*6 +0 Keep in mind that all of the “=” isn’t reallyan equal sign. It’sreally “≡”, “equivalent”, but I don’t have that symbol in an easy type-able form. With large numbers, it’s easier to put the value in question into a calculator and find out the remainder, but then it will be in decimal form, andwe need it in whole numbers. So: Ex 6: 673 mod 17  673/17 = 39.588  17*39 = 663 673-663 =10 673 mod17 =10 Ex 7: 492 mod 16  492/16 = 30.7516*30 = 480  492-480 = 12  492 mod 16 = 12 Try these: 7 mod 5 = 26 mod 3 = 129 mod4 = Decimation Code In decimation code, instead of adding values from a key word, multiply a value that the sender and receiver will know. Ex: Code HEAD SOUTH, code: 3 H E A D S O U T H 7 4 0 3 18 14 20 19 7 x3 21 12 0 9 54 42 60 57 21 26mod2=2 = 16 =8 =5 V M A J C Q I F V Now decode: V M A J C Q I F V 21 12 0 9 2 16 8 5 21 /3 Where valuesare two small to divide by three, keep adding 26 till the value is a multiple of the key 2+26=28 16=42 8+26=34 5=57 +26=54 60 7 4 0 3 18 14 20 19 7 H E A D S O U T H With this code, A willalways beA because 0 multiplied by any number willalwaysbe 0.  Advantages o Simple to code o More difficult to decode  Disadvantages o Does not encode repeat letters o A is always A o Even with a small-valued key, the numbers to code will be large, and they will onlyget larger with bigger keys o The kay can’t be just anything; it has to work when decoding. Some values will be too large to decode with, and others will be confused with similar values. Try decoding: key = 5 R H U U  SymmetricCartography o Same key to code and decode o Private key  AsymmetricCartography o Public key to encode o Private code to decode How is AsymmetricCartography possible Through prime factorization, a method discovered and developed by Rivest, Shamir, and Adleman in 1977; it is known as the RSA publickey encryption scheme. Ex: 24 = 2x2x2x3 Ex: 90 = 2x3x3x5 9x10  9=3x3;10=2x5 Using both modular arithmetic (24=4 mod 20)and prime factorization, they wereable to create a complex coding system. Two prime numbersaremultiplied to get the public key, and then only the decoder knows the prime numbers to factorize by. Ex: 11,483*45,161 = 518,583,763 Ex: 3,130,832,609 =31,91*100,699 Where we are looking at smaller values, they use numbers that aremillions of digits long. Even with a quick computer it would take a long time to find the prime factors of the numbers. Answers: T I K K UK R  FO R W AR D T I K K U K R 19 8 10 10 20 10 17 +26=34 +26=36 36 36 -14 -20 -19 -14 -20 -14 -19 5 14 17 22 0 17 3 F O R W A R D 7 mod 5 =2 26 mod 3 =2 129 mod4 = 1 R H U U;Key = 5  TR E E R H U U 17 7 20 20 +26=43+26=69+26=95 +26=33+26=59+26=85 /5 /5 19 17 4 4 T R E E

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##### ISBN: 9780133951929

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