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| author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
| commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
| tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3886/CH15/EX15.8/15_8.sce | |
| parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
| download | Scilab-TBC-Uploads-master.tar.gz Scilab-TBC-Uploads-master.tar.bz2 Scilab-TBC-Uploads-master.zip | |
Diffstat (limited to '3886/CH15/EX15.8/15_8.sce')
| -rw-r--r-- | 3886/CH15/EX15.8/15_8.sce | 23 |
1 files changed, 23 insertions, 0 deletions
diff --git a/3886/CH15/EX15.8/15_8.sce b/3886/CH15/EX15.8/15_8.sce new file mode 100644 index 000000000..8d8a838f4 --- /dev/null +++ b/3886/CH15/EX15.8/15_8.sce @@ -0,0 +1,23 @@ +//Two incloned planes
+//refer fig.15.10 (a),(b) and (c)
+//Let the assembly move down the 60 degree plane by an acceleration a m/sec^2
+//Consider the block weighing 100 N
+//Applying equilibrium conditions
+N1=50 //N
+mu=1/3
+//From law of friction
+F1=mu*N1 //N
+//T+((100*a)/(9.81))=69.93
+//Now consider 50 N block
+N2=50*cosd(30) //N
+//From the law of friction
+F2=mu*N2
+//((50*a)/(9.81))-T=-39.43
+//Solving we get
+a=(69.93-39.43)*9.81/(100+50) //m/sec^2
+T=69.93-(100*1.9947/9.81) //N
+printf("\na=%.4f m/sec^2\nT=%.2f N",a,T)
+
+
+
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