Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 3 (Q.No. 42)
42.
The resultant of the two forces P and Q is R. If Q is doubled, the new resultant is perpendicular to P. Then
Discussion:
12 comments Page 2 of 2.
Amare said:
5 years ago
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".
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".
Anonymus said:
4 years ago
Thanks @Raj.
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