MOMENT OF A FORCE ABOUT A POINT

Transcription

1 MOMENT OF A FORCE ABOUT A POINT The tendency of a body to rotate about an axis passing through a specific point O when acted upon by a force (sometimes called a torque). 1

2 APPLICATIONS A torque or moment of 12 N m is required to rotate the wheel. Which one of the two grips of the wheel above will require less force to rotate the wheel? 2

3 Couple Moment The couple-moment is known as a free vector, meaning that it can be moved anywhere in space without changing its meaning. 3

4 3.14 Equivalent Couples 135 N 100mm 135 N 135 N 90 N 150mm 90 N 100mm 135 N 100mm 100mm Figure shows three couples which act successively on the same rectangular box. As seen in the preceding section, the only motion a couple can impart to a rigid body is a rotation. Since each of the three couples shown has the same moment M ( same direction and same magnitude M = 135 N. m ), we can expect the three couples to have the same effect on the box. 4

5 MOMENT OF A FORCE ABOUT A POINT Magnitude of a moment Mo = F d N.m Mo = Magnitude of the moment of F around point O d = Perpendicular distance from O to the line of action of F 5

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7 DIRECTION OF MOMENT OF A FORCE Moment produces a rotation. Direction determined by using the Right- Hand Rule. The thumb points along the moment axis and the other fingers are curled following the sense of rotation. Could be clockwise (CW) or anti/counter clockwise (CCW). 7

8 Direction of Moment Choose the convenient sense of rotation for each analysis. 8

9 Calculating moment Scalar Analysis M o = F d M o = F (r sin θ) 9

10 Sample problem 3.1 A 450N vertical force is applied to the end of a lever which is attached to a shaft at O. Determine (a) the moment of the 450N force about O (b) the horizontal force applied at A which creates the same moment about O (c) the smallest force applied at A which creates the same moment about O (d) how far from the shaft a 1100N vertical force must act to create the same moment about O (e) whether any one of the forces obtained in parts b,c and d is equivalent to the original force. 10

11 Solution a) Moment about O The perpendicular distance from O to the line of action of the 450N force is d = (0.6m) cos 60 = 0.3 m The magnitude of the moment about O of the 450N force is Mo = Fd = (450 N) (0.3m) = 135 N.m Since the force tends to rotate the lever clockwise about O, the moment will be represented by a vector Mo perpendicular to the plane of the figure and pointing into the paper.we express this fact by writing. Mo = 135 N.m 11

12 Solution b. Horizontal Force In this case, we have d = (0.6m) sin 60 = 0.52 m Since the moment about O must be 135N.m, we write Mo = Fd 135 N.m = F (0.52 m) F = 260 N F= 260 N 12

13 Solution c. Smallest Force Since Mo = Fd, the smallest value of F occurs when d is maximum. We choose the force perpendicular to OA and note that d = 0.6m, thus Mo = Fd 135 N.m = F (0.6 m) F = 225 N F= 225 N 30 13

14 Solution d N Vertical Force In this case Mo = Fd yields 135 N.m = (1100 N)d d = 0.12 m OB cos 60 = OB = d d cos 60 OB = 0.24 m e. None of the forces considered in parts b,c and d is equivalent to the original 450N force. Although they have the same moment about O, they have different x and y components. In other words, although each forces tends to rotate the shaft in the same manner, each causes the lever to pull on the shaft in a different way. 14

15 Sample problem 3.2 A force of 800 N acts on as bracket as shown. Determine the moment of a force about B. 15

16 Solution F y = 800 sin A F x = 800 cos 60 B M B = xf y + yf x = 0.2 (800 sin 60 ) (800 cos 60 ) = = Nm

17 Sample problems N A 135N force acts on the end of the 0.9 m lever as shown. Determine the moment of the force about O 17

18 Solution 135N The force is replaced by two components, one component P in the direction of OA and one component Q perpendicular to OA. Since O is on the line of action of P, the moment of P about O is zero and the moment of the 135N force reduces to the moment of Q, which is clockwise and, thus, is represented by a negative scalar. Q = (135N) sin 20 = 46.2 N Mo = -Q (0.9m) = -(46.2N)(0.9m) = -41.6N.m Since the value obtained for the scalar Mo is negative, the moment Mo points into the paper. We write Mo = 41.6 N.m 18

19 EXAMPLE 1 Given: A 400 N force is applied to the frame and = 20. Find: The moment of the force at A. 1) Resolve the force along x and y axes. 2) Determine M A using scalar analysis. 19

20 Solution + F y = -400 sin 20 N + F x = -400 cos 20 N + M A = {(400 cos 20 )(2) + (400 sin 20 )(3)} N m = 1160 N m 20

21 Problems 3.1 A 90-N Force is applied to the control rod AB as shown. Knowing that the length of the rod is 225mm, determine the moment of the force about point B by resolving the force into components along AB and in a direction perpendicular to AB. 21

22 Solution 22

23 Problems 3.2 A 90-N Force is applied to the control rod AB as shown. Knowing that the length of the rod is 225mm, determine the moment of the force about point B by resolving the force into horizontal and vertical components. 23

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25 Solution m 4.8 m A 3N force P is applied to the lever which controls the auger of a snow blower. Determine the moment of P about A when αis equal to 30º. 25

26 Equivalent System Sliding VECTOR MOVING A FORCE ON ITS LINE OF ACTION 26

27 MOVING A FORCE OFF OF ITS LINE OF ACTION Free VECTOR 27

28 Resultant of a Force and Couple System 28

29 3.20 Further Reduction of a Force and Couple System If R and M RP are perpendicular to each other, the force-couple system at P can be further reduced to a single resultant force. Will be the case for system consisting either : (a) concurrent force (b) coplanar force (c) parallel forces (a) concurrent force 29

30 (b) Coplanar force systems (c) Parallel force systems 30

31 The moment of the resultant force about the grip the moment of all the forces about the grip F R d = F 1 d 1 + F 2 d 2 + F 3 d 3 31

32 Sample problems N 600N 100N 250N A 4.80m long beam is subjected to the forces shown. Reduce the given system of forces to a) an equivalent force couple system at A b) single force or resultant 32

33 solution use F R d = F 1 d 1 + F 2 d 2 + F 3 d 3 ; 33

34 solution 150N 600N 100N 250N F1 F2 F3 + F R = = N = 600 N + SM A = F 1 d 1 + F 2 d 2 + F 3 d 3 = - 600(1.6m) + 100(2.8m) 250(4.8) = Nm 34

35 solution F R = 600 [N] A d B F R d = SM A (600) d = 1880 Nm d = 1880/600 = 3.13 m Thus, the single force equivalent to the given system; F R = 600 [N], d = 3.13 [m] 35

36 Solution 0.25 m 0.85 m d D 16.4 N 16.4 N 14 N A B C = 2.1 m D L R For equivalence, S M D F : = - R R = 46.8 N : - (0.25)(16.4) (1.10)(16.4) ( d)(14) = - L (46.8) d = 46.8 L»Eq 1 36

37 Example 3 (cont) (a)for d = m, substitute the value of d in Eq 1, (0.625) = 46.8 L L = m The resultant passes through a point m to the right of D (b) For L = 1.05 m, substitute the value of L in Eq. 1, d = 46.8 (1.05) d = m 37

38 THE END 38