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)

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(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.

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation:

hi i’m sorry to call us today so we will get back with y’all and we will have to do it haha omg was the day