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Solutions - Chapter 4 - its good
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b) F : F 120N MPa 2. (A) O O 382 kPa O Divide to convert MPa to kPa An angle of 16 would be required for the block to move down the ramp. The ramp is inclined at 25 c) (steeper angle) and hence will cause the box to slide. An acceleration due to gravity of pushes the box down the ramp at an increasing rate. doutput 3. (D) Friction Force is always parallel to the plane and opposite to the direction whether mm on flat ground or on an incline. :: d 5 m 4. (D) Normal force is always perpendicular (at to the surface. 41. tan 5. (B) O O 0 O Therefore, the largest possible angle before the car starts to slide down is 26. 6. (A) u tang tan 15 Chapter 4 Solutions 7. (A) 1. (B) 8. (A) W Wx F FF 0 N N 0 N. N 300N Wx Wsin0 9. (D) Chapter Solutions 166 b) F : F 120N MPa 2. (A) O O 382 kPa O Divide to convert MPa to kPa An angle of 16 would be required for the block to move down the ramp. The ramp is inclined at 25 c) (steeper angle) and hence will cause the box to slide. An acceleration due to gravity of pushes the box down the ramp at an increasing rate. doutput 3. (D) Friction Force is always parallel to the plane and opposite to the direction whether mm on flat ground or on an incline. :: d 5 m 4. (D) Normal force is always perpendicular (at to the surface. 41. tan 5. (B) O O 0 O Therefore, the largest possible angle before the car starts to slide down is 26. 6. (A) u tang tan 15 Chapter 4 Solutions 7. (A) 1. (B) 8. (A) W Wx F FF 0 N N 0 N. N 300N Wx Wsin0 9. (D) Chapter Solutions 166 10. (C) Weight, drag and lift are the forces no Sum the forces in the vertical direction to find the normal thrust. force in terms of T. 11. (A) Lubricating the 2 surfaces will lower the EF 0 N Tsin15 0 coefficient of friction hence reducing the N 260 Tsin15 frictional force. Sum the forces in the horizontal direction. Sub the 12. frictional force equation into Ff and then substitute the normal equation into N. Factorise T out and solve. P Psin25 25 25 Tcos15 Ff 0 Pcos25 Tcos15 (0) 0 Tcos15 0(260 Tsin15) 0 Tcos15 46 0 0 N T(cos15 0) 46 Sum the forces in the vertical direction and equate to 0 to find the normal force in terms of P. 0 14. N Psin25 0 N Psin25 250 P P Sum the forces in the horizontal direction and equate Psin40 to 0. Sub the frictional force equation into Ffand then Pcos40 un 40 substitute the normal equation into N. Factorise P out and solve. Treat this as a simultaneous equation problem. N FH Pcos25 Ff 0 Sum the forces in the vertical direction to find the normal Pcos25 (0) 0 force in terms of P. Pcos25 0 (Psin25 250) 0 Pcos25 0 62 0 0 P(cos25 0) 62 Psin40 0 N 170 Psin40 62 N Sum the forces in the horizontal direction. Sub the frictional force equation into Ff and then substitute 13. the normal equation into N. Factorise P out and solve. 260N it Ff 0 (0) 0 T 0(170 Psin40) 0 Tsin15 102 0 0 Tcos 15 P(0 cos40) 102 N cos40 167 Chapter 4 Solutions 10. (C) Weight, drag and lift are the forces no Sum the forces in the vertical direction to find the normal thrust. force in terms of T. 11. (A) Lubricating the 2 surfaces will lower the EF 0 N Tsin15 0 coefficient of friction hence reducing the N 260 Tsin15 frictional force. Sum the forces in the horizontal direction. Sub the 12. frictional force equation into Ff and then substitute the normal equation into N. Factorise T out and solve. P Psin25 25 25 Tcos15 Ff 0 Pcos25 Tcos15 (0) 0 Tcos15 0(260 Tsin15) 0 Tcos15 46 0 0 N T(cos15 0) 46 Sum the forces in the vertical direction and equate to 0 to find the normal force in terms of P. 0 14. N Psin25 0 N Psin25 250 P P Sum the forces in the horizontal direction and equate Psin40 to 0. Sub the frictional force equation into Ffand then Pcos40 un 40 substitute the normal equation into N. Factorise P out and solve. Treat this as a simultaneous equation problem. N FH Pcos25 Ff 0 Sum the forces in the vertical direction to find the normal Pcos25 (0) 0 force in terms of P. Pcos25 0 (Psin25 250) 0 Pcos25 0 62 0 0 P(cos25 0) 62 Psin40 0 N 170 Psin40 62 N Sum the forces in the horizontal direction. Sub the frictional force equation into Ff and then substitute 13. the normal equation into N. Factorise P out and solve. 260N it Ff 0 (0) 0 T 0(170 Psin40) 0 Tsin15 102 0 0 Tcos 15 P(0 cos40) 102 N cos40 167 Chapter 4 Solutions EFy 0 F0 N 0 N X cos15 3863 N W 350 N Remember to sub uN into Ff where 0 F Wsin15 (0) 21. F (0 x 3863) : F 2 69 N F Fsin20 F 19. 20 20 Fcos20 Wsin20 N y W The angle is positioned between the Wcos20 and the force vector due to alternate angles. The force vector is broken up into its components shown F above. Apply the sum of forces equation in both the N vertical and horizontal directions to solve for F using simultaneous equations ( substitution). EFy 0 0 400 N 0 N 0 N Fsin20 400 N Wcos20 200cos20 188 N EFH Fcos20 FF 0 F 0 x 188 75 Fcos20 (0) 0 Fcos20 0(Fsin20 400) 0 Fx Fcos20 0 32 0 F F Wsin20 0 F(cos20 0) 32 F F Wsin20 : F 75 200sin20 143 N 32 20. 22. a) W 105N F Wsin40 w 40 EFx Wcosyo 105 Ff 105 1 N un 105 0 40) 169 Chapter 4 Solutions EFy 0 F0 N 0 N X cos15 3863 N W 350 N Remember to sub uN into Ff where 0 F Wsin15 (0) 21. F (0 x 3863) : F 2 69 N F Fsin20 F 19. 20 20 Fcos20 Wsin20 N y W The angle is positioned between the Wcos20 and the force vector due to alternate angles. The force vector is broken up into its components shown F above. Apply the sum of forces equation in both the N vertical and horizontal directions to solve for F using simultaneous equations ( substitution). EFy 0 0 400 N 0 N 0 N Fsin20 400 N Wcos20 200cos20 188 N EFH Fcos20 FF 0 F 0 x 188 75 Fcos20 (0) 0 Fcos20 0(Fsin20 400) 0 Fx Fcos20 0 32 0 F F Wsin20 0 F(cos20 0) 32 F F Wsin20 : F 75 200sin20 143 N 32 20. 22. a) W 105N F Wsin40 w 40 EFx Wcosyo 105 Ff 105 1 N un 105 0 40) 169 Chapter 4 Solutions b) 0 100 x cos40 1501 :N 76 0 55 Ff Wsin40 0 c) Ff 55 N x sin40 N un NUN 64 1u(76) : 0 d) 25. OR tan0 tan 40 y sin10 23. x w Wcoslo FF 4500N 100N 2500 N Fg EFy 0 0 N N Wcos10 X cos10 EF 0 N 984 N N 0 N 4500 N FEH Ff 100 Wsin10 100 X sin10) 2500 Ff 25001 Ff 273 M un NUN 4500 : u 0 24. Sled will be waxed wood on wet snow or rubber on 150N dry snow since they both have lower coefficient of friction than what is needed. Hence, it will have a 55 N smaller frictional force and so sled will slide. N Chapter 4 Solutions 170 b) 0 100 x cos40 1501 :N 76 0 55 Ff Wsin40 0 c) Ff 55 N x sin40 N un NUN 64 1u(76) : 0 d) 25. OR tan0 tan 40 y sin10 23. x w Wcoslo FF 4500N 100N 2500 N Fg EFy 0 0 N N Wcos10 X cos10 EF 0 N 984 N N 0 N 4500 N FEH Ff 100 Wsin10 100 X sin10) 2500 Ff 25001 Ff 273 M un NUN 4500 : u 0 24. Sled will be waxed wood on wet snow or rubber on 150N dry snow since they both have lower coefficient of friction than what is needed. Hence, it will have a 55 N smaller frictional force and so sled will slide. N Chapter 4 Solutions 170 52 sin60 1,20 Apply sum of moments about the foot of the ladder 8 x sin60 6 m to find the reaction force at the wall. We now apply sum of moments about the foot of the EMfoot ladder to find the horizontal distance from the man to w(4) W(1) 0 the base of the ladder. :.F 32 N (250 x 2) 0 (562 x 6) 500 Find the reaction forces at the ground. x EF Use trig to find how far up the ladder the man is. N 150 0 Ff N y Draw a vector triangle to find the resultant force at the ground and its direction. 3 60 R 150 O 32 m R 1 54 N Therefore, he can climb 6 up the ladder before the ladder starts to slip. tano 32 29. F 30. 5m h FF,w Nw Ff Im 3m h N W FFG Find the height of the wall using Pythagoras. 0 0 Chapter 4 Solutions 172 52 sin60 1,20 Apply sum of moments about the foot of the ladder 8 x sin60 6 m to find the reaction force at the wall. We now apply sum of moments about the foot of the EMfoot ladder to find the horizontal distance from the man to w(4) W(1) 0 the base of the ladder. :.F 32 N (250 x 2) 0 (562 x 6) 500 Find the reaction forces at the ground. x EF Use trig to find how far up the ladder the man is. N 150 0 Ff N y Draw a vector triangle to find the resultant force at the ground and its direction. 3 60 R 150 O 32 m R 1 54 N Therefore, he can climb 6 up the ladder before the ladder starts to slip. tano 32 29. F 30. 5m h FF,w Nw Ff Im 3m h N W FFG Find the height of the wall using Pythagoras. 0 0 Chapter 4 Solutions 172 In the above diagram, NG is the normal force at the 31. a) ground and Ff,G is the friction force between ground and ladder. Ff,G is directed to the right since the ladder ny has tendency to slip to the left. Nw is the normal force at the wall and Ff,w is the friction force between wall x and ladder. It is directed upwards as it has tendency T to slip downwards. W represents the 100 N weight D force. Wcos15 w Use Pythagoras to find the height of the wall. Wsin15 32 h2 12 h 2 To convert 80 tonnes to kN, multiply 10. Apply sum of moments about G. 0 L Wcos15 0 L 800 X cos15 Nw (2) 0 L 7 773 kN Using the friction force formula gives us the following b) Use the lift to drag ratio to find drag which relationship. From this, we can then substitute it into use to find the thrust. the above equation to find Nw. Ff,w 0 100(0) w(2) 0(1) 0 .D 88kN 0 T D Wsin15 0 T 48 800sin15 We can now find Ff,w substituting Nw into the below T 255 kN equation. 32. Ff,w 0 x Nw : N ny Ff,g Nw 0 x Ff,g 16 N D Wcos18 w NG 0 NG N Ff,w 0 To convert 12 tonnes to kN, multiply 10. 0 L Wcos18 0 L 120cos18 114 kN 173 Chapter 4 Solutions In the above diagram, NG is the normal force at the 31. a) ground and Ff,G is the friction force between ground and ladder. Ff,G is directed to the right since the ladder ny has tendency to slip to the left. Nw is the normal force at the wall and Ff,w is the friction force between wall x and ladder. It is directed upwards as it has tendency T to slip downwards. W represents the 100 N weight D force. Wcos15 w Use Pythagoras to find the height of the wall. Wsin15 32 h2 12 h 2 To convert 80 tonnes to kN, multiply 10. Apply sum of moments about G. 0 L Wcos15 0 L 800 X cos15 Nw (2) 0 L 7 773 kN Using the friction force formula gives us the following b) Use the lift to drag ratio to find drag which relationship. From this, we can then substitute it into use to find the thrust. the above equation to find Nw. Ff,w 0 100(0) w(2) 0(1) 0 .D 88kN 0 T D Wsin15 0 T 48 800sin15 We can now find Ff,w substituting Nw into the below T 255 kN equation. 32. Ff,w 0 x Nw : N ny Ff,g Nw 0 x Ff,g 16 N D Wcos18 w NG 0 NG N Ff,w 0 To convert 12 tonnes to kN, multiply 10. 0 L Wcos18 0 L 120cos18 114 kN 173 Chapter 4 Solutions 36. Chapter 5 Solutions 1. (D) Work is equal to the sum of the change in potential energy and the change in kinetic energy. Since the trolley is pushed on level L ground (no change in height hence 0 PE) and at a constant speed (no change in speed hence 0 KE), D the total work done will be 0. k w Wcos15 2. (A) Work done the constant force of 10 kN to move the body a distance of 50 m up the slope is: 0 L Wcos15 0 W 10 x 50 500 kJ L 1020cos15 985 kN Work done due to potential energy (mgh) only since 0 there is no change in speed hence no change in KE: T D Wsin15 0 T 40 1020sin15 304 kN W APE 37. W mghf mghi W 33 x 15) 0 W 4501 kJ 0 The work done the constant force is converted to PE. The energy loss due to heat will hence be 50kJ. 5714 Fstabiliser : Fstabiliser 714 k N 3. (A) W 180 x 50 J :W 9 38. Find friction force produced the braking pads impacting the brake disc. Multiply friction force 4. (B) 2 due to having 2 normal forces applied to the : P 667 W brake pads. 5. (B) Inclined road suggests an increase in height NUN hence PE increases. Accelerating suggests an F 0 x x 2 3040 N increase in speed hence KE increases. This friction force acts at a tangent to the rotary 6. (D) Barbara lifts a total of 200kg which accounts motion of the wheel slowing it down. to a weight of 3040 N mass 20x10 200 kg Fweight 200 x 10 N She produces a power of: Multiply frictional force the distance to the centre to find its moment. 40 W M Fd M 3040 x 0 516 Nm 175 Chapter Solutions 36. Chapter 5 Solutions 1. (D) Work is equal to the sum of the change in potential energy and the change in kinetic energy. Since the trolley is pushed on level L ground (no change in height hence 0 PE) and at a constant speed (no change in speed hence 0 KE), D the total work done will be 0. k w Wcos15 2. (A) Work done the constant force of 10 kN to move the body a distance of 50 m up the slope is: 0 L Wcos15 0 W 10 x 50 500 kJ L 1020cos15 985 kN Work done due to potential energy (mgh) only since 0 there is no change in speed hence no change in KE: T D Wsin15 0 T 40 1020sin15 304 kN W APE 37. W mghf mghi W 33 x 15) 0 W 4501 kJ 0 The work done the constant force is converted to PE. The energy loss due to heat will hence be 50kJ. 5714 Fstabiliser : Fstabiliser 714 k N 3. (A) W 180 x 50 J :W 9 38. Find friction force produced the braking pads impacting the brake disc. Multiply friction force 4. (B) 2 due to having 2 normal forces applied to the : P 667 W brake pads. 5. (B) Inclined road suggests an increase in height NUN hence PE increases. Accelerating suggests an F 0 x x 2 3040 N increase in speed hence KE increases. This friction force acts at a tangent to the rotary 6. (D) Barbara lifts a total of 200kg which accounts motion of the wheel slowing it down. to a weight of 3040 N mass 20x10 200 kg Fweight 200 x 10 N She produces a power of: Multiply frictional force the distance to the centre to find its moment. 40 W M Fd M 3040 x 0 516 Nm 175 Chapter Solutions
Solutions - Chapter 4 - its good
Subject: Engineering Studies
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