Question 1206728



given:

first force, {{{F[1] = 160N}}} at {{{45}}}⁰
second force, {{{F[2] = 280N}}} at {{{0}}}⁰

The vertical component of the forces is calculated as:

{{{F[y]=160*sin(45)+280*sin(0)}}}

{{{F[y]=160*(1/sqrt(2))+280*(0)}}}

{{{F[y]=160/sqrt(2)}}}

{{{F[y]=113.14}}}


The horizontal component of the forces is calculated as:

{{{F[x]=160*cos(45)+280*cos(0)}}}

{{{F[x]=160*(1/sqrt(2))+280*(1)}}}

{{{F[x]=160/sqrt(2)+280}}}

{{{F[x]=393.14}}}


The direction of the forces is:

{{{theta=tan^-1(F[y]/F[x])}}}

{{{theta=tan^-1(113.14/393.14)}}}

{{{theta=tan^-1(0.28778552169710536)}}}

{{{theta=0.28021352584028169}}} radians 

{{{theta=16.06}}}° 


Magnitude of resultant force:

{{{R=sqrt(113.14^2+393.14^2)}}}

{{{R=409.096}}}