##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
l=2.##units of length
a=math.sqrt(3.)##degree
b=30.##dgree
c=60.##degree
v=1.##length
Pc=1.154##tonf compressive
##CALCULATIONS
R=(v*l)/a##tonf
D=math.sqrt((R)**2+(v)**2)##tonf
T=41.##degree
P=l*math.cos(b)##tonf tensile
Pa=Pc*math.cos(b)##tonf tensile
p=(l*math.cos(b))/((1/2.)+(Pc))/(1/2.)##tonf compressive
##RESULTS
print'%s %.2f %s'%('the resolving horizontally = ',p,' tonf compressive')
#due to round off error answer is different
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
v=3.##tonf
p1=6.0##tonf,compressive
p2=5.19##tonf,tensile
a=30.##degree
b=60.##degree
p3=7.##tonf,compressive
##CALCULATIONS
P1=p2*math.sin(b)##tonf,tensile
P2=1./2.*P1##tonf,compressive
P3=p1*math.cos(a)-p3*math.cos(b)##tonf,compressive
P4=P1*math.cos(a)*math.sqrt(3.)/P3##tonf,acting towards the left
R=P1*math.sin(a)##tonf,acting downwards
D=math.sqrt((P4)**2+(R)**2)##tonf
T=(R/P4)##to the horizantal
##RESULTS
print'%s %.2f %s'%('the direction reaction= ',D,' to the horizantal')
print'%s %.2f %s'%('the direction reaction = ',T,' to horizantal')
#converted into degrees
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
R1=5.##tonf
R2=7.##tonf
P=5.77##tonf,compressive
m=11.56##tonf
a=30.##degree
##CALCULATIONS
P=-(0.15+4.5084+1+1-10)##tonf
P1=math.sqrt(P)
##RESULTS
print'%s %.2f %s'%('the methods of sections in the force= ',-P1,' tonf')
#converted into degrees