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
"m_ha = F_{lift} - m_hg-T, \\;\\; T = m_f(g+a) \\Longrightarrow m_ha = F_{lift} - m_hg-m_f(g+a)."
a) Therefore, "F_{lift} = (m_h+m_f)(g+a) = (7650\\,\\mathrm{kg}+1250\\,\\mathrm{kg})(9.8\\,\\mathrm{m\/s^2}+0.8\\,\\mathrm{m\/s^2}) = 94340\\,\\mathrm{N}."
b) The tension in the cable is "T = m_f(g+a) = 1250\\,\\mathrm{kg}\\cdot(9.8\\,\\mathrm{m\/s^2}+0.8\\,\\mathrm{m\/s^2}) = 13250\\,\\mathrm{N}."
c) The tension is equal to the tension, so it is also "13250\\,\\mathrm{N}."
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