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

`F_p = < - √(3) , -3 >\\\\F_o = < √(3) , -1 >`

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 = < (1)/(2) , (√(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 ( (\pi )/(6) ) , - cos ( (\pi )/(3) ) >\\\\u_p = < (√(3) )/(2) , -(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 = < (1)/(2) , (√(3) )/(2) > [ < 0 , - 4 > . < (1)/(2) , (√(3) )/(2) > ]\\\\F_p = < (1)/(2) , (√(3) )/(2) > [ -2√(3) ]\\\\F_p = < -√(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 = < (√(3) )/(2) , - (1)/(2) > [ < 0 , - 4 > . < (√(3) )/(2) , - (1)/(2) > ] \\\\F_o = < (√(3) )/(2) , - (1)/(2) > [ 2 ] \\\\F_o = < √(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 > = < √(3), -1 > + < -√(3), -3 > \\\\< 0 , - 4 > = < √(3) - √(3) , -1 - 3 > \\\\< 0 , - 4 > = < 0 , - 4 >` .. proven