Question

Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise from the positive y axis. The second force has magnitude F2=46.1 N and is pointing at an angle of θ2=28.8 counterclockwise from the negative x axis.
a) What is the x component of the net force acting on the object?

b) What is the y component of the net force acting on the object?

c) What is the x component of the equilibrant?

d) What is the y component of the equilibrant?

e) What is the magnitude of the equilibrant?

f) What is the angle the equilibrant makes with the x axis? You can use either degrees (enter unit as deg) or radians (enter unit as rad) in the range from 0 to 360 deg (or 0 to 2 rad).

Answer 1

Answer:

a)Fnx = -26.92 N

b)Fny= 8.4 N

c)Fex = 26.92 N

d)Fey = - 8.4 N

e)β = 18.34° clockwise from the positive x axis

Explanation:

Look at the attached graphic

a) What is the x component of the net force acting on the object?

Fnx:x component of the net force acting on the object

Fnx= F₁x+ F₂x

F₁x=  33.4*sin 23.8° = 13.48 N

F₂x= - 46.1*cos 28.8°= - 40.4 N

Fnx =  13.48 N- 40.4 N= -26.92 N

b) What is the y component of the net force acting on the object?

Fny= F₁y+ F₂y

F₁y=  33.4*cos 23.8° =30.6 N

F₂y= - 46.1*sin 28.8°= -22.2 N

Fny=30.6 N-22.2 N = 8.4 N

c) What is the x component of the equilibrant?

Fex: component of the equilibrant

Fex = - Fnx

Fex = - ( -26.92 N)

Fex = 26.92 N

d) What is the y component of the equilibrant?

Fey = - Fnx

Fey = - ( 8.4 N)

Fey = - 8.4 N

e) What is the magnitude of the equilibrant?

$F_{e}$ :  equilibrant force

$F_{e} = \sqrt{(F_{e}x)^{2}+{(F_{e}y)^{2} }$

$F_{e} = \sqrt{(26.92)^{2}+{(-8.4)^{2} }$

$F_{e} = 28.2 N$

f) What is the angle the equilibrant makes with the x axis?

$\beta =tan^{-1} (\frac{-8.94}{26.96} )$

β = -18.34° or β = 18.34° clockwise from the positive x axis