Question

the resultant of two. force P and Q acting at an angle equal to​

Answer 1

In case I :

∣

∣

∣

∣

P

+

Q

∣

∣

∣

∣

=

P

2

+Q

2

+2Pθcosθ

⇒(2MH)

P

2

+Q

2

=

P

2

+Q

2

+2PQcosθ

⇒(2M+1)

2

(P

2

+Q

2

)=P

2

+Q

2

+2PQcosθ

⇒

2(PQ)

(4m

2

+4m)

(P

2

+Q

2

)=cosθ

In case II

∣

∣

∣

P

2

+Q

2

∣

∣

∣

=

P

2

+Q

2

+2PQsinθ

⇒(2m−1)

P

2

+Q

2

=

P

2

+Q

2

+2PQsinθ

⇒(2m−1)

2

(P

2

+Q

2

)=P

2

+Q

2

+2PQsinθ

⇒

2(PQ)

(4m

2

−4m)

(P

2

+Q

2

)=sinθ

So, tanθ=

cosθ

sinθ

=

(2PQ)

4m

2

−4m

(4m

2

+4m)

(2PQ)

(P

2

+Q

2

)

(P

2

+Q

2

)

⇒tanθ=

(4m)

(4m)

(m+1)

(m−1)

=

m+1

m−1

Here Q=α So, tanα=

m+1

m−1