1.
Fnet = F1+F2+F3 = (-2.45 i -2.40j) + (5.40 i) + (43.5 i + 47.0 j) = (46.45i + 50.0j) N
a) θ=tan^-1(Fnety/Fnetx)= tan^-1(50.0/46.45) = 47.11 deg …above x axis
b) Fnet = sqrt(Fnetx^2+Fnety^2)= sqrt(46.45^2+50.0^2) = 68.25 N
m= Fnet/anet = 68.25/3.50 = 19.5 kg
c) vf= vi+at = 0 + 3.50*18.0 = 63.0 m/s
d) vfx = vix + anetx*t = 0 + 3.50cos47.11*18.0 = 42.88 m/s i
vfy = viy + anety*t = 0 + 3.50sin47.11*18.0 = 46.16 m/s j
vf = (42.88 m/s) i + (46.16 m/s) j
2.
Applying Newton’s second law at point B,
T3=mg = 500 N
Now applying Newton’s second law at point A horizontally,
T2cos27 - T1cos50 =0
T2= T1*(cos50/cos27) -------(1)
Now applying Newton’s second law at point A vertically,
T2sin27 + T1sin50 = T3----------(2)
From (1),
[T1*(cos50/cos27)]sin27 + T1sin50 = T3
T1*(cos50*tan27) + T1sin50 = T3
T1[(cos50*tan27) + sin50] = T3
T1[(cos50*tan27) + sin50] = 500
T1= 457.22 N
Three forces acting on an object are given F_1 = (-2.45i - 2.40j)N, F_2 = (5.40i)N,...
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