CE Statics Lecture 10

FORCE SYSTEM RESULTANTS So far, we know that for a particle to be in equilibrium, the resultant of the force system acting on the particle is to be equal to zero. That is:  F = 0 In a later stage, we are going to see that  F = 0 is required for a particle to be in equilibrium, but it is not sufficient to make that particle in equilibrium since we know that forces could create momentum which will tend to rotate the particle. In the coming lectures, moments about a point or axis will be discussed.

CROSS PRODUCT If we have two vectors A and B, then the CROSS PRODUCT of the two vectors is: C = A  B (C is equal to A cross B) The magnitude of C is: C = AB sin  The direction of C is perpendicular to the plane containing A and B. Then C = A  B = ( AB sin  ) u C A B C uCuC  AB sin 

Laws of Operation (1) A  B  B  A BUT A  B = - B  A A B C = A  B  A B - C = A  B 

Laws of Operation (2) Multiplying by a Scalar a (A  B) = (a A)  B = A  (aB) = (A  B) a (3) A  (B + D) = (A  B) + (A  D)

Cartesian Vector Formulation Use C = AB sin (  ) to find the cross product of two vectors. i  j = k; i  k = -j ; i  i = 0 j  k = i; j  i = -k; j  j = 0 k  i = j; k  j = -i; k  k = 0 i  j = ij sin 90 = (1)(1)(1) = 1 i  i = ii sin (0) = 0

If A and B are two vectors, then: A  B = (Ax i + Ay j + Az k)  (Bx i + By j + Bz k) = AxBx (i  i) +AxBy (i  j) + AxBz (i  k) + AyBx (j  i) +AyBy (j  j) + AyBz (j  k) + AzBx (k  i) +AzBy (k  j) + AzBz (k  k) = (AyBz – AzBy) i – (AxBz – AzBx) j + (AxBy – AyBx) k This equation can be written as: i j k A  B = Ax Ay Az Bx By Bz This is nothing but a determinant whose first raw consists of unit vectors i, j, and k and its second and third raws represent the x, y, and z components of the two vectors A and B.

MOMENT OF A FORCE – SCALAR FORMATION A moment of a force about a point or axis is an indication or measure of the tendency of the force to rotate the particle about that point or axis. Fx is perpendicular to the handle of the wench and located a distance (dy) from point (O). Fx tends to turn the pipe about the z-axis. The turning effect will be larger if Fx or dy gets larger. This turning effect is sometimes referred to as TORQUE, but most frequently it is called MOMENT (M 0 )z y x z O Fx dy

If Fz is applied to the wrench at dy from O. Fz will tend to rotate the pipe about the x-axis and (M O )x is produced. The force Fz and distance dy lie in the (x-z) plane, which is perpendicular to the x-axis ( the moment axis) y x z O Fz dy

If Fy is applied to the wrench, no moment will be produced at point (O) since the line of action of Fy passes through point (O). y x z O Fy dy

Generally, the moment of force F about point (O) which lies in a plane perpendicular to the moment axis, has a magnitude and direction. Therefore, the moment is a vector quantity. Magnitude M O = F  d Where d is called the moment arm, which is perpendicular distance from (O) to the line of action of force F. F d O

Direction F d O F d O MOMO MOMO

Resultant Moment of a System of Coplannar Forces F1, F2, and F3 lie in the (x-y) plane. Their distances from point (O) are d1, d2, and d3, respectively. + M O =  F  d y x z F1 F2 F3 d1 d2 d3 O

Assignment , 54, 57, 60, 64, 65