Let us determine the forces that are applied to the helicopter, namely the forces of gravity, lift force, tension. We use the projection on the vertical axis and get
mha=Flift−mhg−T,T=mf(g+a)⟹mha=Flift−mhg−mf(g+a).
a) Therefore, Flift=(mh+mf)(g+a)=(7650kg+1250kg)(9.8m/s2+0.8m/s2)=94340N.
b) The tension in the cable is T=mf(g+a)=1250kg⋅(9.8m/s2+0.8m/s2)=13250N.
c) The tension is equal to the tension, so it is also 13250N.
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