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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 = 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 g = 9.8 m/s²). • The normal force N (directly up). • 10 kg and is being pulled The force of static friction fs (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 fs = μs N), 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 cosμs N = 0 • F sin N-mg = 0 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) F = ΤΕ ΑΣΦ ?

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 = 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 g = 9.8 m/s²). • The normal force N (directly up). • 10 kg and is being pulled The force of static friction fs (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 fs = μs N), 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 cosμs N = 0 • F sin N-mg = 0 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) F = ΤΕ ΑΣΦ ?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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 = 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 g = 9.8 m/s²). • The normal force N (directly up). • 10 kg and is being pulled The force of static friction fs (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 fs = μs N), 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 cosμs N = 0 • F sin N-mg = 0 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) F = ΤΕ 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 = 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 g = 9.8 m/s²). • The normal force N (directly up). • 10 kg and is being pulled The force of static friction fs (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 fs = μs N), 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 cosμs N = 0 • F sin N-mg = 0 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) F = ΤΕ ΑΣΦ ?
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