A model airplane has momentum given by p=[(-0.75kg.m/s3)t2 + (3.0kg.m/s)] i + (0.25kg.m/s2)t j. Find the components Fx, Fy, and Fz of the net force on the airplane.

Respuesta :

Answer:

[tex]F_x[/tex] = -1.5t

[tex]F_y[/tex] = 0.25

[tex]F_{z}[/tex] = 0

Explanation:

Given equation;

p = [(-0.75 kgm/s³)t² + (3.0 kgm/s)] i + (0.25 kgm/s²)t j.

From Newton's law, the rate of change of momentum of a body is the net force acting on that body. i.e

∑F = [tex]\frac{dp}{dt}[/tex]       -----------(i)

Substitute the equation of p into equation (i) and differentiate with respect to t as follows;

∑F = [tex]\frac{dp}{dt}[/tex] = [tex]\frac{d| [(-0.75)t^{2} + (3.0)] i + (0.25)t j|}{dt}[/tex]

∑F = [tex]\frac{dp}{dt}[/tex] = [tex][-1.5t + 0]i + 0.25j[/tex]

∑F = [tex][-1.5t + 0]i + 0.25j[/tex]

But

∑F = [tex]F_xi + F_yj + F_zk[/tex]

Where;

[tex]F_x, F_y, F_z[/tex] are the components of the net force in the x, y and z direction respectively.

=> [tex]F_xi + F_yj + F_zk[/tex] =  [tex][-1.5t + 0]i + 0.25j[/tex] = [tex]-1.5ti + 0.25j[/tex]

=> [tex]F_x[/tex] = -1.5t

=> [tex]F_y[/tex] = 0.25

=> [tex]F_{z}[/tex] = 0