Answer:
The magnitude will be "353.5 N". A further solution is given below.
Explanation:
The given values is:
F = 500 N
According to the question,
In ΔABC,
⇒ [tex]\angle BCA = (90-30)[/tex]
⇒ [tex]=60^{\circ}[/tex]
then,
⇒ [tex]\angle BAC=(180-45-60)[/tex]
⇒ [tex]=75^{\circ}[/tex]
Now,
The corresponding angle will be:
⇒ [tex]\angle FAC=60^{\circ}[/tex]
⇒ [tex]\angle FAB=70+60[/tex]
⇒ [tex]=135^{\circ}[/tex]
Aspect of F across the AC arm will be:
= [tex]F\times cos(60)[/tex]
On putting the values of F, we get
= [tex]500\times (.5)[/tex]
= [tex]200 \ Newton[/tex]
Component F along the AC (in magnitude) will be:
= [tex]F\times cos(135)[/tex]
= [tex]500\times (-.707)[/tex]
= [tex]-353.5 \ N \[/tex]
