What is the resultant force on the airplane?

Question

An airplane's engines provide a thrust of 15,000 N due south. It flies in a wind that pushes it 3,000 N Northwest at an angle of 160 degrees measured from East. What is the resultant factor on the airplane? Give the magnitude and direction.

Answer 1

The plane is flying south, but it gets pushed in the northwest direction with the wind coming from the east.  What we need to do is draw the vectors of the plane's direction.  We will have two vectors: thrust vector and wind vector.  These vectors will broken down by their x and y components.

 

Thrust vector:

 

F_tx = 0 N

F_ty = -15000 N

 

 

Wind vector:

 

This vector is northwest.  This direction lies in the second quadrant in a coordinate system and 160 degrees lies in the quadrant.

 

F_wx = 3000cos(160) = -2819.08 N

F_wy = 3000sin(160) = 1026.06 N

 

 

To find the resultant vector of the airplane, we need to combine all the x components and combine all the y components.  Then use the Pythagorean theorem.

 

R_x = F_tx + F_wx = 0 N - 2819.08 N = -2819.08 N

R_y = F_ty + F_wy = -15000 N + 1026.06 N = -13973.94 N

 

R = √(R_x² + R_y²)

 

 

To get the direction of the vector, we use the inverse tangent.

 

θ = tan^-1(R_y / R_x)

 

 

If R_x is negative and R_y is negative.  Therefore, the resultant vector will lie in the 3rd quadrant.