Mechanical Engineering - Engineering Mechanics - Discussion

Given condition: two forces are equal, then P=Q,

R^2=P^2+Q^2+2PQcosθ.

Substitute P=Q.
R^2=P^2+P^2+2P^2cosθ.

Take common 2P^2 term,
R^2=2P^2(1+csoѲ).

From trigonometry..1+cosθ=2cos^2 θ/2.

So substitute this value above then,
R^2=2P^2(2cos^2 θ/2).
R^2=4P^2(cos^2 θ/2).

Cancel the squares on both sides by applying a root.

Then,
The resultant R=2Pcos Ѳ/2.

R^2=A^2+B^2-2ABcos(θ)

According to Cosine Law OR Parallelogram law, θ is the Angle between forces A,& B.

Now A=B=P The given condition and take square root both sides.
R = (P^2+P^2-2*P*P*cos(θ))^0.5
R = (2P^2-2P^2cos(θ))^0.5.

Take Common 2P^2.
R=(2P^2(1-cos(θ))^0.5 ------> (1).

According to Trigonometry.

((1-cos(θ))/2)^0.5 = cos(θ/2).
So, divide and multiply (1) with 2.
R=(2*2P^2(1-cos(θ))/2)^0.5.

Separate the power.
R=(2*2*P^2)^.05 * ((1-cos(θ))/2)^0.5
((1-cos(θ))/2)^0.5 = cos(θ/2).
R=2P*cos(θ/2)

@Ajaypal.

How can we replace as you mentioned in equation A?

Please explain briefly.

Here, Cos 2θ= cos^2 (θ) - sin^2 (θ).

R^2 = P^2+Q^2+2PQcosθ.
R^2 = P^2+P^2+2P^2cosθ.
or
R^2 = 2P^2(1+cosθ)----> Eq <A>

Trig. Identity Cos 2θ= cos^θ-sin^θ
So cosθ= cos^θ/2-sin^θ/2,
Putting above in Eq A and replacing 1-sin^2 θ/2 with cos^2 θ/2,
You will get the answer.

Can you explain clearly?

We know that,
R2 = P2+Q2+2PQcos(a).
R2 = 2P2 (1+cos(a)).
R2 = 2P2 (2cos2(a/2)).
R = 2Pcos(a/2).