Mathematics, 18.06.2020 21:57 tohanab

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.

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Answer from: ayoismeisalex

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

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

Answer from: Quest

b. 24 mm

step-by-step explanation:

use the pythagorean theorem.

a^2 + b^2 = c^2

a^2 + 7^2 = 25^2

a^2 + 49 = 625

a^2 = 576

a = 24

answer: b. 24 mm

What is the length of the missing leg in this right triangle ?

Answer from: Quest

answer: im gonna go with milli

step-by-step explanation:

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