1. Follow the figure for Problem 3.6 Using a Crowbar to Remove a Nail
Moment about a Point
70° 
18 in. A B C
Follow the figure for Problem 3.6 Using a Crowbar to Remove a Nail
Copyright: Geraldine Milano, PE

2. This is the picture. Here is some information ...
70° 
18 in. A B C 4”

3. A nail is being removed from a board at pt. C.
70° B C
Copyright: Geraldine Milano, PE

4. A vertical force of 200 lb is needed to lift the nail.
70°
200 # B C 4”

5. The crowbar is supported against the board at pt. B.
70° B C 4”

6. Copyright: Geraldine Milano, PE
A ‘pulling’ force at A will ‘rotate’ the crowbar about the pivot point, B. A  P
70° B C
Copyright: Geraldine Milano, PE

7. Watch the rotation about pt. B
70° 
18 in. A B C
Copyright: Geraldine Milano, PE

8. Watch the rotation about pt. B
80° 
18 in. A B C
Copyright: Geraldine Milano, PE

9. Watch the rotation about pt. B
90° 
18 in. A B C
Copyright: Geraldine Milano, PE

10. Watch the rotation about pt. B
100° 
18 in. A B C
Copyright: Geraldine Milano, PE

11. Watch the rotation about pt. B
110° 
18 in. A B C
Copyright: Geraldine Milano, PE

12. Watch the rotation about pt. B
100° 
18 in. A B C
Copyright: Geraldine Milano, PE

13. Watch the rotation about pt. B
90° 
18 in. A B C
Copyright: Geraldine Milano, PE

14. Watch the rotation about pt. B
80° 
18 in. A B C
Copyright: Geraldine Milano, PE

15. Watch the rotation about pt. B
70° 
18 in. A B C
Copyright: Geraldine Milano, PE

16. Show the components of P
70° 
18 in. A B C P P
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17. P ‘pushes’ along the member
70° 
18 in. A B C P P
Copyright: Geraldine Milano, PE

18. Copyright: Geraldine Milano, PE
P will ‘rotate’ the bar P
70° 
18 in. A B C P
Copyright: Geraldine Milano, PE

19. Copyright: Geraldine Milano, PE
Therefore, P x 18” = MB P
18 in. A  P
70° B C
Copyright: Geraldine Milano, PE

20. But, MB = 200# x 4” = 800 in.lb. P A  P 18 in. 70° 800 in.lb. 200# BC 4”

21. Copyright: Geraldine Milano, PE
Therefore, 800 in.lb. = P x 18” P
18 in. A  P
70°
800 in.lb. B C
Copyright: Geraldine Milano, PE

22. Now work on the geometry.P
70° 
18 in. A B C
Given:  = 10°
70° P
Opposite interior angles are equal = 70° P
Copyright: Geraldine Milano, PE

23. Find the angle between P and P
Given:  = 10°
20° A
10°
70° P
18 in.
The complement of 70° = 20° P
70° B C
Copyright: Geraldine Milano, PE

24. Copyright: Geraldine Milano, PE
P = P cos 30° P
20° A
10°
70° P
18 in.
70° B C
Copyright: Geraldine Milano, PE

25. Copyright: Geraldine Milano, PE
 800 in.lb. = P cos 30° x 18” P
18 in. A
30° P
70°
800 in.lb. B C
Copyright: Geraldine Milano, PE

26. Copyright: Geraldine Milano, PE
800 in.lb. = P cos 30° x 18” in.lb. = P
(18”cos 30°)
P = lb.
Copyright: Geraldine Milano, PE

27. Copyright: Geraldine Milano, PE
P = lb. P
20° A
10°
70° P
18 in. P
70° B C
Copyright: Geraldine Milano, PE

28. Copyright: Geraldine Milano, PE
Now for the last part… what is the minimum force required to produce the same moment ?
Copyright: Geraldine Milano, PE

29. What is the minimum force, P?
70°
 = ?
18 in. A B C P 4”

30. Copyright: Geraldine Milano, PE
Since, 800 in.lb. = P x 18” P
18 in. A  P
70°
800 in.lb. B C
Copyright: Geraldine Milano, PE

31. Copyright: Geraldine Milano, PE
Pmin = 800 in.lb./18” = 44.4 lb. P
18 in. A
70°
800 in.lb. B C
Copyright: Geraldine Milano, PE

32. Copyright: Geraldine Milano, PE
So… … the minimum effort results when the force is perpendicular to the member , … ie; force is perpendicular to the moment arm.
Copyright: Geraldine Milano, PE