CHE 210 HW2 Solution - Force 2 direction (degrees): -20...

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>> type pin % Steve Swinnea % Program to find resultant force on a single point % with three 2D forces acting on it. F1m = input('Force 1 magnitude (N): '); F1d = input('Force 2 direction (degrees): '); F2m = input('Force 2 magnitude (N): '); F2d = input('Force 2 direction (degrees): '); F3m = input('Force 3 magnitude (N): '); F3d = input('Force 3 direction (degrees): '); %Calculations: F1 = [F1m*cosd(F1d) F1m*sind(F1d)]; F2 = [F2m*cosd(F2d) F2m*sind(F2d)]; F3 = [F3m*cosd(F3d) F3m*sind(F3d)]; Fres = F1+F2+F3; Resm = norm( Fres ); Resd = atan2( Fres(2) , Fres(1) )*180/pi; %must use atan2 not atand fprintf('\nMagnitude of Resultant: %0.1f\n',Resm); fprintf('Direction of Resultant: %0.1f\n',Resd); >> pin Force 1 magnitude (N): 400
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Unformatted text preview: Force 2 direction (degrees): -20 Force 2 magnitude (N): 500 Force 2 direction (degrees): 30 Force 3 magnitude (N): 700 Force 3 direction (degrees): 143 Magnitude of Resultant: 590.0 Direction of Resultant: 64.9 >> type work %Steve Swinnea %A Program that calculates work from a force %and displacement vector F1 = input('Enter force vector 1: '); F2 = input('Enter force vector 2: '); d = input('Enter displacement vector: '); Res = F1+F2; W = Res*d'; %Assumes both are row vectors Angle = acosd( W/norm( Res )/norm( d ) ); fprintf('\nWork Done: %0.2f\n', W); fprintf('Angle: %0.1f\n', Angle); >> work Enter force vector 1: [15 20 5] Enter force vector 2: [20 -5 7] Enter displacement vector: [10 5 1] Work Done: 437.00 Angle: 12.8...
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This note was uploaded on 09/26/2010 for the course CHE 210 taught by Professor Swinnea during the Fall '08 term at University of Texas at Austin.

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CHE 210 HW2 Solution - Force 2 direction (degrees): -20...

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