Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 58.0 ° counterclockwise from the + x ‑axis, whereas the second force has x ‑ and y ‑components ( − 19.0 N , 16.5 N ) . Express the magnitude and direction of the net force.

magnitude of the net force is 32.49N

and the direction of the net force is 107.91°

Let the first force be F₁

Let the second force be F₂

Let the net force be F

From the question;

(a) F₁ has 17.0N in magnitude and has direction θ = 58.0° counterclockwise from +x axis. i.e

|F₁| = 17.0N

θ = 58.0°

Re-write this force in vector form as follows;

F₁ = |F₁| cos θ i + |F₁| sin θ j

F₁ = [17.0 cos 58.0°] i + [17.0 sin 58.0] j

F₁ = [17.0 x 0.5299] i + [17.0 x 0.8480] j

F₁ = [9.0083] i + [14.416] j

Put the unit as follows;

F₁ = [9.0083 N] i + [14.416 N] j

(b) F₂ is already written in vector form but can be re-written as follows;

F₂ = [-19.0N ] i + [16.5N] j

(c) The net force (F) is the algebraic sum of the two forces F₁ and F₂ as follows;

F = F₁ + F₂

Substitute the values of F₁ and F₂ into the equation above;

F =  [9.0083 N] i + [14.416 N] j + [-19.0N ] i + [16.5N] j

Now, collect like terms;

F =  [9.0083 N] i + [-19.0 N] i + [14.416N ] j + [16.5N] j

F = -9.9917 N i  + 30.916 N j                     ----------------(a)

===============================================================

(a) Now let's calculate the magnitude, |F|, of the net force as follows;

|F| =

|F| =

|F| =

|F| = 32.49N

Therefore, the magnitude of the net force is 32.49N

(b) The direction, θ, is calculated as follows;

tan θ = (y component / x component)

tan θ = (30.916 / -9.9917)

tan θ = (-3.094)

θ = tan ⁻¹ (-3.094)

θ = -72.09°

The negative sign shows that it is measured with respect to the -x axis.

From equation (a), it is shown that the net force is located in the -i and +j directions measured counterclockwise from the +x axis. Therefore, to get the true angle of the net force relative to the positive x axis and counterclockwise, we add 180° to the angle calculated above. i.e

true angle = 180° + (-72.09°)

true angle = 107.91°

Therefore, the direction of the net force is 107.91° counterclockwise from the +x axis.

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

false br