Force x has a magnitude of 1260 pounds, and force y has a magnitude of 1530 pounds. they act on a single point at an angle of 45 degrees to each other. find the magnitude of the equilibrant and the angle between the equilibrant and the 1260-pound force.

Fe= 2579.68 P

α= 24.8°

Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

FX = 1260 P , horizontal (-x)

FY = 1530  P , forming 45° with positive x axis

x-y components FY

FYx= - 1530*cos(45)° = - 1081.87 P

FYy= -  1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

Fny=FYy

Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the  equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny=  -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

Fe= 2579.68 P

Calculation of the direction of  equilibrant force (α)

α= 24.8°

Look at the attached graphic

its c i just checked it