Question

Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° 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.

Answer 1

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the  the second force(F₂)

F₂x= −19.0 N

F₂y=   16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

$F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }$

$F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }$

$F_{n} = 30.11N$

Direction of the net force (β)

$\beta =tan^{-1} (\frac{29.13}{7.62} )$

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

Answer 2

We want to get the total force acting on an object, we will get that by directly adding the vectors of each force. We will get that the magnitude is 30.11N, and the direction is θ = -75.3°

We know that the first force has a magnitude of 17 N and it's oriented 48° counterclockwise, then the components are:

$F_x = 17N*cos(48\°) = 11.38 N\\\\F_y = 17N*sin(48\°) = 12.63 N$

Then the vector that represents this force is:

$F_1 = (11.38N, 12.63N)$

And the other force is defined by the vector:

$F_2 = (-19N, 16.5N)$

To get the net force we just add the two forces:

$F = F_1 + F_2 = (11.38N, 12.63N) + (-19N, 16.5N) = (-7.62N, 29.13N)$

Now we have the net force, the magnitude is just:

$F = (-7.62N, 29.13N)\\\\|F| = \sqrt{(-7.62N)^2 + ( 29.13N)^2} = 30.11N$

To get the direction, we remember that the two components are orthogonal, thus these can represent the catheti of a right triangle where the hypotenuse is equal to the magnitude of the vector.

Then the angle, where the x-component is the adjacent cathetus, is given by.

tan(θ) = 29.13N/-7.62N

θ = Atan(29.13N/-7.62N) = -75.3°

This angle defines the direction of the net force.

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