Force Systems and Equilibrium

Force Systems and Equilibrium - 4 Force systems and...

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4 Force systems and equilibrium 4.1 Addition of forces Force has been defined in Chapter 3 in the context of Newton’s laws of motion. The action of a single force has been quantified by the changes it produces in the motion of a particle. As we have shown that force is a vector quantity, any two forces acting at a Doint may be replaced by a resultant force R (Fig. 4.1). If a third force is now introduced, this may be added to R in just the same way as F2 was added to F, (Fig. 4.2). specific point on that line. The difference between FIA FIB is characterised by the separation d of the lines of action. 4.2 Moment of force By definition, the magnitude of the moment of F (Fig. 4.4) about 0 is Fd. Figure 4.1 Figure 4.4 Clearly any force F acting tangentially to a sphere, radius d, gives a moment of the same magnitude, but the effect is uniquely defined if we associate with the magnitude a direction perpen- dicular to the plane containing F r in a sense given by the right-hand screw rule. The moment of a force may therefore be regarded as a vector with a magnitude Fd = Frsina and in a direction e as defined in ~i~. 4.4. H~~~~ we may write the moment of the force F about the point 0 as (4.1) Figure 4.2 It is obvious that the position of the point of application is important. Consider two forces, equal in magnitude and direction, acting on their effects are clearly not the same. If, now, the force at A is applied at C, the overall effect is not 4-3 altered; however, the internal effects will be The vector Or Cross Product of two vectors A different. B is written AXB and is defined to have a We conclude that the overall effect is governed magnitude IA I IBl sins, where a is the angle by the line of action of the force and not by any between the two vectors. The direction of A X is given by the right-hand screw rule as shown in Fig. 4.5. Note that the vector product is not commutative since by definition x A = -A x B, see Fig. 4.5. If are expressed in terms of their Cartesian components, different points on a body as shown in Fig. 4.3; Mo = Frsinae = Fde Vector Product of two vectors Figure 4.3
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38 Force systems and equilibrium Figure 4.6 i.e. the moment of the resultant of F1 and F2 is equal to the vector sum of the moments of the components. If a force F is replaced by its Cartesian components then the moment about 0 is, by inspection of Fig. 4.7, Mo = (yF,-zFy)i+(zF,-xF,)j + (XF,-yF,)k (4.5) Figure 4.5 A X B = (A,i+Aj+A,k) X (B,i+ Byj+ B,k) We must first consider the vector product of orthogonal unit vectors. By inspection, ixj =k= -jxi jxk=i = -kxj kxi=j = -ixk also hence ixi = jxj = kx k = 0 AxB =A,Byixj+A,B,ixk +Ay Bxj x i+ A, Bj x k + A , Bx k X i + A , By k X j -Ay B, k + A, B,i +A, BjO- A, Byi Figure 4.7 and this is seen to be the same as the vector-algebra definition = A, By k -A, Bj = (AyB,-A,By)i+(A,B,-A,Bz)j M,=rxF=(xi+yj+zk) X (F,i+ Fyj+ F,k) + (AxBy- AyBx)k (44 jk This result is summarised by the following 1 determinant: XY z - 1 - AxB= A, (4.3) Fx Fy Fz Bx By Bz 4.5 Couple From Fig.
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Force Systems and Equilibrium - 4 Force systems and...

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