Question

A 13500 kg jet airplane is flying through some high winds. At some point in time, the airplane is pointing due north, while the wind is blowing from the north and east. If the force on the plane from the jet engines is 35700 N due north, and the force from the wind is 15300 N in a direction 80.0° south of west, what will be the magnitude and direction of the plane's acceleration at that moment? Enter the direction of the acceleration as an angle measured from due west (positive for clockwise, negative for counter-clockwise). a) Magnitude of acceleration = ______ m/s2 b) Direction of acceleration = _______ ° (degrees)

Answer 1

Answer

given,

mass of the jet airplane = 13500 kg

Force on the plane = 35700 N due north

force from wind  = 15300 N in direction 80.0° south of west.

Force = $35700 \vec{j} N$

force by wind = $15300(-cos \theta \vec{i}-sin \theta \vec{j})$ N

                       = $15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j})$ N

net force on the jet airplane(ma)

          $m a = 35700 \vec{j} + 15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j})$

          $\vec{a} = \dfrac{35700}{13500} \vec{j} + \dfrac{15300}{13500}(-cos 80^0 \vec{i}-sin 80^0 \vec{j})$

          $\vec{a} = 2.64\vec{j} -0.197 \vec{i} - 1.116 sin 80^0 \vec{j})$

           $\vec{a} = -0.197 \vec{i} + 1.524 \vec{j}$

$a = \sqrt{-0.197^2+1.524^2}$

a = 1.54 m/s²

$\theta = tan^{-1}(\dfrac{-1.524}{0.197})$

$\theta = -82.63^0$