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Follow the figure for Problem 3.6 Using a Crowbar to Remove a Nail

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Presentation on theme: "Follow the figure for Problem 3.6 Using a Crowbar to Remove a Nail"— Presentation transcript:

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 Copyright: Geraldine Milano, PE

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# B
C 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


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