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Help with 3 questions, I have included teachers feedback aswell from last attempt Assignment 7 Unit B Module 1, Lesson 4: Free-Body Diagrams and Net
Help with 3 questions, I have included teachers feedback aswell from last attempt
Assignment 7 Unit B Module 1, Lesson 4: Free-Body Diagrams and Net Force Lesson 1 ofthis assignment is worth 9 marks. The value of each question is noted in the left margin in parentheses. Remember: all net force problems should include a freebody diagram. 1. Beta convinces Alpha to drag their ice-fishing shack to a new location by pulling it with ropes. Alpha applies a force of 150 N [175\"] on one rope and Beta applies a force of 175 N [190\"] on another rope. What is the net force applied to the shack by Alpha and Beta? Assume the ice is frictionless. Show all work. 2. The next day, Alpha decides to move the shack back to its original location. Alpha (having had a nutritious breakfast) is able to pull the 1800 kg shack at a constant velocity with a force of 190 N. Calculate the coefficient of friction between the shack and the ice. Show all work. Unit B Module 1, Lesson 5: Solving Net Force Problems Lesson 2 ofthis assignment is worth 33 marks. The value of each question is noted in the left margin in parentheses. Remember: all net force problems should include a free-body diagram. 3. A textbook with a mass of2.18 kg is pulled across a table with a horizontal force of 5.5 N [E]. If the table has a coefficient of friction of 0.15, what is the acceleration of the textbook? 1. Beta convinces Alpha to drag their ice-fishing shack to a new location by pulling it with ropes. Alpha applies a force of 150 N [175 ] on one rope and Beta applies a force of 175 N [190 ] on another rope. What is the net force applied to the shack by Alpha and Beta? Assume the ice is frictionless. Show all work. (6) Alpha force : 2/6 Sin ( 1750 ) = app your : sin ( 175 " ) X15ON app : - 11.5 N ( negative because it's in the opposite direction to positive *-axis ) X X cos ( 1750 ) = ads / hyp Start by drawing your free body diagram adj = cos (175' ) x 150N and resolving vector diagram. adj = - 138. IN ( negative as it's app . to positus younis ) Alpha : horizontal veshawl 71.5 8 - 135. 7N Belai I see you have tried to find resolving vectors here but your calculations are y sin ( 190 0 ) = app / hyp incorrect. You are also missing the final steps. " Pp = sin ( 190 " ) * 175RX See Lesson 4, P6, SC7 "Free-Body Cop = . 104. 3 N X Diagrams in 2D" X cas ( 190 ) = adj/ nyp adj = cos( 190-) * 17SN adj = 170- 9 N X Bela horizont + verbal is -104. 3N @ ADLC All Rights reserved ( 2019 ) and 170. 9 N2 of 8 2 Physics 20: Mrs. Tyler Assignment 7 2. The next day, Alpha decides to move the shack back to its original location. Alpha (having had a nutritious breakfast) is able to pull the 1800 kg shack at a constant velocity with a force of 190 N. Calculate the coefficient of friction between the shack and the ice. Show all work. 2.5 (3) normal force = 18 wky + 9. 8 lm / s applied fore = 190 mass = 1800 kg = 17, 658 N gravity = 9.81 m/ s 2 coefficient of Frickin = 190N / 17, 658 N =0.0108 Please show free body diagram and direction for full marks. See Lesson 4, P5 "Free-Body Diagrams and Net Force" After you have completed all questions in Lesson 4, please place your assignment in a safe place until you have finished Lesson 5. Unit B Module 1, Lesson 5: Solving Net Force Problems Lesson 2 of this assignment is worth 33 marks. The value of each question is noted in the left margin in parentheses. Remember: all net force problems should include a free-body diagram. 3. A textbook with a mass of 2.18 kg is pulled across a table with a horizontal force of 5.5 N [E]. If the table has a coefficient of friction of 0.15, what is the acceleration of the textbook? (2) mass = 2. 18kg applies force = 5.5 N (6 ] 1.5/2 coefficient of frichon = 0.15 gravity - 5.8limbs normal face= 2.18kg4 9.81 mls - 21. 37N force of frichu = 0. 15+ 21.37N = 3. aUN For full marks you need to show direction, as well as a 5. SN[E] - 321 0 FBD and indicate direction in your answer. = 2. 2 9 N CEJ a= 2. 2 AN CC] / 2.18 kg a = 1. 0530# 587m/5 APLC All Rights Reserved (2019)Step by Step Solution
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