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author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
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committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3886/CH3/EX3.9 | |
parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
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Diffstat (limited to '3886/CH3/EX3.9')
-rw-r--r-- | 3886/CH3/EX3.9/3_9.txt | 7 | ||||
-rw-r--r-- | 3886/CH3/EX3.9/Ex3_9.sce | 13 |
2 files changed, 20 insertions, 0 deletions
diff --git a/3886/CH3/EX3.9/3_9.txt b/3886/CH3/EX3.9/3_9.txt new file mode 100644 index 000000000..a90e43daf --- /dev/null +++ b/3886/CH3/EX3.9/3_9.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_9.sce', -1)
+Equilibriant is equal and opposite to resultant.
+R=116.52 kN
+alpha=76.82 degree
+x=1.517 m
+As shown in fig.3.18 (a)
\ No newline at end of file diff --git a/3886/CH3/EX3.9/Ex3_9.sce b/3886/CH3/EX3.9/Ex3_9.sce new file mode 100644 index 000000000..7d2158b17 --- /dev/null +++ b/3886/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,13 @@ +//determine equilibriant
+//two 40 kN forces have no moment about the pulley centre hence can be considered acting at pulley centre
+//Accordingly
+Rx=20*cosd(45)-30*cosd(60)-50*cosd(30)+40*cosd(20)-40*sind(30) //kN (towards left)
+Ry=-20*sind(45)-20+20-30*sind(60)-50*sind(30)-40*sind(20)-40*cosd(30) //kN (Downwards)
+R=sqrt(Rx^2+Ry^2) //kN
+alpha=atand(Ry/Rx) //degree
+//Taking moment about A
+MA=20*4-20*4+30*6*sind(60)+50*2*sind(30)-50*2*cosd(30)+40*3*cosd(20)-40*3*sind(30)
+//assume that the resultant intersects AB at a distance x from A,then
+x=MA/Ry //m
+printf("Equilibriant is equal and opposite to resultant.\nR=%.2f kN\nalpha=%.2f degree\nx=%.3f m\nAs shown in fig.3.18 (a)",R,alpha,-x)
+
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