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

To find the magnitude and direction of the plane's acceleration, we need to consider the forces acting on it. The force from the jet engines is 35700 N due north, while the force from the wind is 15300 N in a direction 80.0° south of west. By resolving the forces into their components and adding them, we can calculate the magnitude and direction of the acceleration.

Step-by-step explanation:

To find the magnitude and direction of the plane's acceleration, we need to consider the forces acting on it. The force from the jet engines is 35700 N due north, while the force from the wind is 15300 N in a direction 80.0° south of west.

1. Resolve the force from the jet engines into its north and west components. The north component is 35700 N and the west component is 0 N.
2. Resolve the force from the wind into its north and west components. Using trigonometry, we find that the north component is 15300*cos(80°) N and the west component is -15300*sin(80°) N.
3. Add the north and west components of both forces to find the net north and west components of the total force acting on the plane.
4. The net north component is 35700 N + 15300*cos(80°) N = 35700 N + (-3057 N) = 32643 N.
   The net west component is 0 N + (-15300*sin(80°)) N = -15300 N*sin(80°) = -15170 N.
5. Use the net north and west components to find the magnitude and direction of the plane's acceleration using the formula:
   Magnitude of acceleration = sqrt((net north component)^2 + (net west component)^2)
   Direction of acceleration = arctan((net west component)/(net north component))

Answer 2

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} = (35700)/(13500) \vec{j} + (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 = √(-0.197^2+1.524^2)`

a = 1.54 m/s²

`\theta = tan^(-1)((-1.524)/(0.197))`

`\theta = -82.63^0`