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. What is the magnitude of the equilibrant? Unable to interpret units. Computer reads units as "deg". Help: Physical_Units Tries 3/10 Previous Tries
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:

Fe= 28.2 N : Magnitude of the equilibrant (Fe)

β = 18.34° , clockwise from the positive x axis

Explanation:

Concept of the equilibrant

It is called equilibrant  to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero

To solve this problem we decompose the forces given into x-y components to find the resulting force:

Look at the attached graphic

F₁= 33.4 N  , θ₁=23.8° clockwise from the positive y axis (y+)

F₁x= 33.4 *sin23.8° = 13.48 N

F₁y= 33.4 *cos23.8° =30.6 N

F₂=46.1 N ,  θ₂=28.8 counterclockwise from the negative x axis (x-)

F₂x= -46.1 *cos28.8° = -40.4 N

F₂y=  -46.1 *sin28.8° =  -22.2 N

Components of the resultant in x-y R(x,y)

Rx= 13.48 N -40.4 N = - 26.92 N

Ry= 30.6 N  -22.2 N =  + 8.4 N

Components of the equilibrant in x-y Fe(x,y)

Fex= +26.92 N

Fey=  - 8.4 N

Magnitude of the equilibrant (Fe)

$F_{e} = \sqrt{(F_{ex})^{2}+{(F_{ey})^{2} }$

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

Fe= 28.2 N

Angle the equilibrant makes with the x axis ( β)

$\beta = tan^{-1} (\frac{F_{ey} }{F_{ex} } )$

$\beta = tan^{-1} (\frac-8.4 }{26.92 } )$

β = -18.34°                  

β = 18.34° , clockwise from the positive x axis