ME256Vector Components Vector Addition 3D PDF

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ME 2560 Statics Vector Components and Vector Addition (3D) Cartesian Components of Vectors (3D) o Given the magnitude of a vector and the direction of the vector relative to a set of three reference axes, the vector can be expressed in terms of its components along those axes. o For our convenience, it is beneficial to have the reference axes be a right-handed, mutually perpendicular set. o V , V , and V represent the components of the vector V along the right-handed, mutually perpendicular X, Y, and Z axes. By vector addition, V  V  V  V o The triangles formed by V and V , by V and V , and by V and V are all right triangles. o If the magnitude of the vector V is | V | V , and if the angles that V makes with the X, Y, and Z axes are  x ,  y , and  z , respectively, then V V  V V  Vx  V cos( )  k

 V cos( x ) 



 V cos( x ) i

j

o The unit vectors i , j , and k indicate the positive X, Y, and Z coordinate directions. o The vector u  cos( ) i  cos( ) j  cos( )k is a unit vector in the direction of V . o The angles  x ,  y , and  z are not independent. It can be shown that, cos2  x   cos2  y   cos2  z   1

o The magnitude of V is V  Vx2  Vy2  Vz 2

Kamman – ME 2560: page 1/4

Cartesian Components – Polar and Elevation Angles o The Cartesian components of a vector can also be given in terms of polar and elevation angles. Unlike the three angles relative to the X, Y, and Z axes, these two angles are independent. o In the diagram, the angle  is the polar angle, and the angle  is the elevation angle. o In this case, the Cartesian components are found using a two-step process. First, break V into two components, one in the XY plane and one perpendicular to it (along the Z axis). Then, break the component in the XY plane into two components, one along the X axis and one along the Y axis.





V  V cos( ) cos( ) i  sin( ) j  V sin( )k  V cos( )cos( )



 V cos( )cos( ) i

j

Example 1: Given: A force F has magnitude F  100 (lb) and angles x  40 (deg) ,  y  70 (deg) . Find: Express the force F in terms of the unit vectors i ,

j , and k . Solution: o We know that: cos2  x   cos2  y   cos2 z   1 . So, we can find  z as follows:

 z  cos1



1  cos2  x   cos2  y 



 cos1



1  cos2  40  cos2  70  57.027  57.0 (deg)



F  100 cos(40) i  100 cos(70) j  100 cos(57.027)k

o  76.6  34.2

o Check: | F |  76.62  34.22  54.4 2  99.9828 100 (lb) Kamman – ME 2560: page 2/4

Example 2: (polar and elevation angles) Given: A force F has magnitude F  100 (lb) and angles   24 (deg) ,   33 (deg) . Find: Express the force F in terms of the unit vectors i , j , and k . Solution: o Fx  F cos( )cos( )  100cos(33)cos(24)  76.6164  76.6 (lbs) o Fy  F cos( )sin( )  100cos(33)sin(24)  34.1118  34.1 (lbs) o Fz  F cos( )  100sin(33)  54.4639  54.5 (lbs) o

F  76.6 i  34.1 j  54.5 k

o Check: | F |  76.62  34.12  54.52  100.003  100 (lb)

Example 3: Given: A force F

70 i 50 j 80 k (lb) .

Find: The magnitude and direction of F relative to the X, Y, and Z axes. Solution: o

F  F  70 2  502 802 117.473  117 (lb)

o  x  cos 1  Fx F   cos 1 70 117.473   53.425  53.4(deg) o  y  cos 1  Fy F   cos 1  50 117.473  64.81  64.8(deg) o  z  cos1  Fz F   cos1  80 117.473 132.92 133(deg) o Check:

cos2 53.43  cos2  64.8  cos2 132.9  0.9997  1

Kamman – ME 2560: page 3/4

Vector Addition using Cartesian Components (3D) To add two or more vectors, simply express them in terms of the same unit vectors, and then add like components. As before, we call the sum of the vectors the resultant. Example 4: Given:

F  100 i  175 j  200 k (lb) F  75 i  25 j 100 k (lb) F  120 i 100 j  300 k (lb)

Find: The magnitude and the direction of the resultant force F . Solution: o

F  100  75 120  i  175  25 100  j  200 100 300  k  95  100 j  20

o | F |  F  952  1002  2002  242.951  243 (lb) o  x  cos1  Fx F   cos1  95 242.951 113.018 113(deg) o  y  cos1  Fy F   cos 1 100 242.951  65.6942  65.7 (deg) o  z  cos1  Fz F   cos 1  200 242.951 145.408 145(deg) o Check:

cos2 113.018  cos2  65.6942  cos2  145.408  1

Kamman – ME 2560: page 4/4...