#import emoji: square

= Practice Problems Ch 8-9-10

#square.black = Initially got wrong

1. ✅ B 
2. ✅ B 
3. ✅ B 
4. ✅ D
5. ✅ A
6. ✅ A
7. ✅ B
8. ✅ C (higher Young's modulus means less strain)
9. ✅ D
10. ✅ C
11. ✅ B
12. ✅ A
13. ✅ A
14. ✅ B
15. Net force and torque must be 0
    1. ✅ 1068.2
    2. ✅ 709.43
17. ❌ 8.46 m
18. #square.black $f_s = F_W$, where $F_W$ is the normal force of the wall on the ladder \
  This can be found by setting the bottom of the ladder as the pivot point and making sure the torques cancel out
  $F_W = 622$ \
  Normal force of floor on ladder: $F_N = 874 + 400$ (has to cancel out the weight of the ladder and the man) \
  $f_s = mu F_N arrow mu = f_s / F_N = 0.49$
19. ✅ C
20. ✅ Left is 370 N, right is 130 N
23. ✅ A

26. ✅ $F/A = Y ((Delta L)/L) arrow m g/A = Y ((Delta L)/L) arrow m = (Y A Delta L)/(L g) = 11.54 approx 12 "kg"$

31. ✅ A
32. ✅ C

37. ✅ D
38. ✅ A

50.
  1. ✅ 200 N
  2. ✅ 102

62. ✅ B

65. ✅ D (0)

67. #square.black A

76. ✅ E

105. ✅ A
