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Part E - Solving for Two Variables Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying a Expression Primers. Enter the answer as two numbers (either fraction or decimal), separated by a comma, with C first. ► View Available Hint(s) C, D= ΨΕΙ ΑΣΦ ?

Part E - Solving for Two Variables Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying a Expression Primers. Enter the answer as two numbers (either fraction or decimal), separated by a comma, with C first. ► View Available Hint(s) C, D= ΨΕΙ ΑΣΦ ?

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Part E - Solving for Two Variables Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers. Enter the answer as two numbers (either fraction or decimal), separated by a comma, with C first. ► View Available Hint(s) C, D = Submit ΠΫΠΙ ΑΣΦ ?
Transcribed Image Text:Part E - Solving for Two Variables Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers. Enter the answer as two numbers (either fraction or decimal), separated by a comma, with C first. ► View Available Hint(s) C, D = Submit ΠΫΠΙ ΑΣΦ ?
Part D - Isolating a Variable with a Coefficient In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have 3C + 4D = 5 2C+ 5D = 2 ● ● None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process? D = ²/3 - ²/C -³/C C = 1-³/D C=-D 3 ○ D = O Submit 5 Previous Answers Correct
Transcribed Image Text:Part D - Isolating a Variable with a Coefficient In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have 3C + 4D = 5 2C+ 5D = 2 ● ● None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process? D = ²/3 - ²/C -³/C C = 1-³/D C=-D 3 ○ D = O Submit 5 Previous Answers Correct
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Part F - Example: Finding Two Forces (Part I) Two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in (Figure 1) has a mass m = 10 kg and is being pulled by a force F on a table with coefficient of static friction μs 0.3. Four forces act on it: • The applied force F (directed 0 = 30° above the horizontal). • The force of gravity Fg = mg (directly down, where 9 = • The normal force N (directly up). • The force of static friction f, (directly left, opposing any potential motion). If we want to find the size of the force necessary to just barely overcome static friction (in which case ƒ = µN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: F cos 0 μs N=0 F sin 0+N-mg = 0 F = In order to find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for Fin terms of m, g, 0, and μs (no N). Express your answer in terms of m, g, 0, and μs. View Available Hint(s) = 9.8 m/s²). ΠΫΠΙ ΑΣΦ ?
Transcribed Image Text:Part F - Example: Finding Two Forces (Part I) Two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in (Figure 1) has a mass m = 10 kg and is being pulled by a force F on a table with coefficient of static friction μs 0.3. Four forces act on it: • The applied force F (directed 0 = 30° above the horizontal). • The force of gravity Fg = mg (directly down, where 9 = • The normal force N (directly up). • The force of static friction f, (directly left, opposing any potential motion). If we want to find the size of the force necessary to just barely overcome static friction (in which case ƒ = µN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: F cos 0 μs N=0 F sin 0+N-mg = 0 F = In order to find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for Fin terms of m, g, 0, and μs (no N). Express your answer in terms of m, g, 0, and μs. View Available Hint(s) = 9.8 m/s²). ΠΫΠΙ ΑΣΦ ?
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