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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3792/CH5/EX5.15/Ex5_15.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3792/CH5/EX5.15/Ex5_15.sce')
| -rw-r--r-- | 3792/CH5/EX5.15/Ex5_15.sce | 26 |
1 files changed, 26 insertions, 0 deletions
diff --git a/3792/CH5/EX5.15/Ex5_15.sce b/3792/CH5/EX5.15/Ex5_15.sce new file mode 100644 index 000000000..9e7b29fc3 --- /dev/null +++ b/3792/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,26 @@ +// SAMPLE PROBLEM 5/15
+clc;funcprot(0);
+// Given data
+N=1500;// rev/min
+theta_1=60;// degree
+r=5/12;// ft
+ABbar=14/12;// ft
+
+// Calculation
+omega=(2*%pi*N)/60;// rad/s
+a_B=r*omega^2;// ft/sec^2
+omega_AB=29.5;// rad/sec
+a_AB_n=ABbar*omega_AB^2;
+// If we adopt an algebraic solution using the geometry of the acceleration polygon, we first compute the angle between AB and the horizontal. With the law of sines, this angle becomes 18.02 degree.
+theta_2=18.02;// degree
+function[X]=acceleration(y)
+ X(1)=((a_B*cosd(theta_1))+(a_AB_n*cosd(theta_2))-(y(2)*sind(theta_2)))-y(1);
+ X(2)=((a_B*sind(theta_1))-(a_AB_n*sind(theta_2))-(y(2)*cosd(theta_2)))-0;
+endfunction
+y=[1000 1000];
+z=fsolve(y,acceleration)
+a_AB_t=z(2);// ft/sec^2
+a_A=z(1);// ft/sec^2
+r=ABbar;// ft
+alpha_AB=a_AB_t/r;// rad/sec^2
+printf("\nThe acceleration of the piston A,a_A=%4.0f ft/sec^2 \nThe angular acceleration of the connecting rod AB,alpha_AB=%4.0f rad/sec^2",a_A,alpha_AB);
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