Question

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 4 right angle0,−4 in the directions parallel to and normal to the plane that makes an angle of StartFraction pi Over 3 EndFraction π 3 with the positive​ x-axis. Show that the total force is the sum of the two component forces.

Answer 1

Answer:

$F_p = < - \sqrt{3} , -3 >\\\\F_o = < \sqrt{3} , -1 >$

Step-by-step explanation:

- A plane is oriented in a Cartesian coordinate system such that it makes an angle of ( π / 3 ) with the positive x - axis.

- A force ( F ) is directed along the y-axis as a vector < 0 , - 4 >

- We are to determine the the components of force ( F ) parallel and normal to the defined plane.

- We will denote two unit vectors: ( $u_p$ ) parallel to plane and ( $u_o$ ) orthogonal to the defined plane. We will define the two unit vectors in ( x - y ) plane as follows:

- The unit vector ( $u_p$ ) parallel to the defined plane makes an angle of ( 30° ) with the positive y-axis and an angle of ( π / 3 = 60° ) with the x-axis. We will find the projection of the vector onto the x and y axes as follows:

                         $u_o$ = < cos ( 60° ) , cos ( 30° ) >

                         $u_o = < \frac{1}{2} , \frac{\sqrt{3} }{2} >$

- Similarly, the unit vector ( $u_o$ ) orthogonal to plane makes an angle of ( π / 3 ) with the positive x - axis and angle of ( π / 6 ) with the y-axis in negative direction. We will find the projection of the vector onto the x and y axes as follows:

                        $u_p = < cos ( \frac{\pi }{6} ) , - cos ( \frac{\pi }{3} ) >\\\\u_p = < \frac{\sqrt{3} }{2} , -\frac{1}{2} >$

- To find the projection of force ( F ) along and normal to the plane we will apply the dot product formulation:

- The Force vector parallel to the plane ( $F_p$ ) would be:

                          $F_p = u_p(F . u_p)\\\\F_p = < \frac{1}{2} , \frac{\sqrt{3} }{2} > [ < 0 , - 4 > . < \frac{1}{2} , \frac{\sqrt{3} }{2} > ]\\\\F_p = < \frac{1}{2} , \frac{\sqrt{3} }{2} > [ -2\sqrt{3} ]\\\\F_p = < -\sqrt{3} , -3 >$

- Similarly, to find the projection of force ( $F_o$ ) normal to the plane we again employ the dot product formulation with normal unit vector (  $u_o$  ) as follows:

                         $F_o = u_o ( F . u_o )\\\\F_o = < \frac{\sqrt{3} }{2} , - \frac{1}{2} > [ < 0 , - 4 > . < \frac{\sqrt{3} }{2} , - \frac{1}{2} > ] \\\\F_o = < \frac{\sqrt{3} }{2} , - \frac{1}{2} > [ 2 ] \\\\F_o = < \sqrt{3} , - 1 >$

- To prove that the projected forces ( $F_o$ ) and ( $F_p$ ) are correct we will apply the vector summation of the two orthogonal vector which must equal to the original vector < 0 , - 4 >

                       $F = F_o + F_p\\\\< 0 , - 4 > = < \sqrt{3}, -1 > + < -\sqrt{3}, -3 > \\\\< 0 , - 4 > = < \sqrt{3} - \sqrt{3} , -1 - 3 > \\\\< 0 , - 4 > = < 0 , - 4 >$  .. proven