Mechanical Engineering - Engineering Mechanics - Discussion

Thanks @Raj.

Tan a= Qsin0/P + Qcos0.

When Q is double, then.
Tan90 = 2Qsin0/P + 2Qcos0.

We know tan90 = infinity.
P + 2Qcos0 = zero.
P = -2Qcos0.
R = root P^2 + Q^2 + 2PQCOS0.
Putting P = -2Qcos0.
R = Q.

For P and Q.

R^2 = P^2 + Q^2 + 2PQcosθ -----> (1).

For P and 2Q (when Q is doubled).
W^2= P^2+ (2Q) ^2 +2P (2Q) cosθ.
W^2= P^2+4Q^2 +4PQcosθ ------> (2) and,
When Q is doubled the new resultant "W" and "P" are perpendicular. Then from Pythagoras theorem.

W^2+P^2= (2Q) ^2 ----------------> (3).

Next substituting equation (2) into equation (3) we will get;
P^2+2PQcosθ=0 -----------------> (4).

Finally substituting equation (4) into equation (1) we will have.
R=Q.

Then the answer is "B".

Thank You @Raj.

Thanks @Popy.

Resultant of two forces P and Q is R.

R = P + Q.
|R| = √(P^2 + Q^2 + 2PQCosθ).

When Q is doubled new resultant R\' is;
R = P + 2Q,
Since it is perpendicular to P, dot product of R\' and P is zero,
(R).(P) = 0,
(P + 2Q).(P) = 0,
P^2 + 2PQCosθ = 0.

(Dot product of P and P is P^2 and that of P and 2Q is 2PQCosθ)
Cosθ = -P/(2Q).
R = √(P^2 + Q^2 + 2PQCosθ),
R = √(P^2 + Q^2 + 2PQ*(-P/(2Q)),
R = √(P^2 + Q^2 - P^2),
R = Q.

P=Q should be the correct answer.

Tan90 =1/0.

Good question, thanks for giving this.

Thank you all for the given explanation.

@Jahnanarayan.

Any number divided by zero it give infinite value.