AP EAPCET 08-Jul-2022 Shift 2 Paper

Forces given are as shown

Resloving co,ponents we have,

For equilibrium ;
ΣFy=0
F2‌sin‌60°+F1‌sin‌30°=F3‌cos‌45°
⇒F2×

√3

2

+F1×

1

2

=F3×

1

√2

⇒√3F2+F1=√2F3...(i)
and ΣFx=0
F2‌cos‌60°=F3‌sin‌45°+F1‌cos‌30°
⇒F2×

1

2

=F3×

1

√2

+F1×

√3

2

⇒F2=√2F3+√3F1...(ii)

Subtracting Eq.(ii) from Eq.(i) we get,
√3F2+F1−F2=−√3F1
⇒(√3−1)F2=−F1(1+√3)
⇒F2=F1(

1+√3

1−√3

)
⇒F2=−

F1(√3+1)2

(√3−1)(√3+1)

⇒F2=

−F1(3+1+2√3)

3−1

or
F2=−F1(2+√3)
Now, putting the value of F2 in Eq.(i), we get
√3[−F1(2+√3)]+F1=√2F3
F1[−2√3−3+1]=√2F3
F3=√6−√2