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By Dan B. Marghitu

Advanced Dynamics: Analytical and Numerical Calculations with MATLAB offers an intensive, rigorous presentation of kinematics and dynamics whereas utilizing MATLAB as an built-in instrument to resolve difficulties. subject matters offered are defined completely and directly,allowing basic rules to emerge via purposes from parts equivalent to multibody platforms, robotics, spacecraft and layout of advanced mechanical units. This publication differs from others in that it makes use of symbolic MATLAB for either thought and functions. specific recognition is given to recommendations which are solved analytically and numerically utilizing MATLAB. The illustrations and figures generated with MATLAB make stronger visible studying whereas an abundance of examples supply extra aid.

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C) Find the angle between the vectors V1 and V2 . Numerical application: |V1 | = V1 = 3 , |V2 | = V2 = 3, a = 4, b = 5, and c = 3. Solution (a) The vectors V1 and V2 are given by rBG rBG = V1 , |rBG | rBG rBP rBP = V2 V2 = V2x ı + V2y j + V2zk = |V2 | . |rBP | rBP V1 = V1x ı + V1y j + V1zk = |V1 | Next the vectors rBG and rBP will be calculated. From Fig. 16, the coordinates of the points B, D, P, and Q are B = B(xB , yB , zB ) = B(0, b, 0) = B(0, 5, 0), G = G(xG , yG , zG ) = G(a, 0, c) = G(4, 0, 3), and P = P(xP , yP , zP ) = P(a, b/2, 0) = P(4, 5/2, 0).

Solution The magnitudes, Vi , of the vectors Vi and the coordinates, xi , yi , zi , of the points Ai , i = 1, 2, 3, 4 are introduced with MATLAB as V(1)=10; x(1)= 3; x(2)= 5; x(3)=-4; x(4)= 4; V(2)=25; y(1)= 1; y(2)=-3; y(3)=-3; y(4)= 2; V(3)=15; z(1)= 7; z(2)= 4; z(3)= 1; z(4)=-3; V(4)=40; % magnitudes V_i % A_1 % A_2 % A_3 % A_4. 11 Examples 43 and the x, y, z components of the vectors Vi are Vix = Vi cos θix , Viy = Vi cos θiy , Viz = Vi cos θiz . To calculate the direction cosines and components of the vectors for i = 1, 2, 3, 4 the MATLAB statement for var=startval:step:endval, statement end is used.

In Fig. 13a, the rectangular component of the vector F on the OA direction is f, with the magnitude |f| = f . The vector F acts at an angle β with the positive direction of the x-axis. Find the magnitude |F| = F of the vector F. Numerical application: f =20, α = 30◦ , and β = 60◦ . b 20 18 16 14 F y 12 F y a 10 A θ f 8 6 f 4 α β O Fig. 11 Examples 37 Solution The component of F on the OA direction is |F| cos θ = f . From Fig. 13, the angle θ of the vector F with the OA direction is θ = β − α = 60◦ −30◦ = 30◦.

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