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