two forces P and Q acting at a point at an angle theta have their resultant (2n+1)√P^2+Q^2 and when at an angle (90 degree- thetA)their resultant is (2n+1)√P^2+Q^2
show that tan theta=(n-1)÷(n+1)
Answers
Explanation:
OK. Let's work in two dimensions i⃗ and j⃗ , it would save much time since we are working with two vectors.
For simplicity's sake, I'm going to assume Q⃗ acts in the i⃗ direction, and P⃗ acts in a direction at an angle θ to Q⃗ . Note that this θ is the angle between the i⃗ axis and the vector P⃗ .
So, resolving P⃗ and Q⃗ into their respective components when the angle between them is θ , we get,
P⃗ =Pcosθi⃗ +Psinθj⃗
Q⃗ =Qi⃗
Of course you know that P=∥∥P⃗ ∥∥ and Q=∥∥Q⃗ ∥∥ .
Let's call the resultant of these vectors Rθ→ .
Rθ→=P⃗ +Q⃗ =[Pcosθ+Q]i⃗ +Psinθj⃗
∥∥∥Rθ→∥∥∥=[Pcosθ+Q]2+[Psinθ]2−−−−−−−−−−−−−−−−−−−−√
From your question, ∥∥∥Rθ→∥∥∥ is,
∥∥∥Rθ→∥∥∥=(2n+1)P2+Q2−−−−−−−√
Equating the ∥∥∥Rθ→∥∥∥ 's,
[Pcosθ+Q]2+[Psinθ]2−−−−−−−−−−−−−−−−−−−−√=(2n+1)P2+Q2−−−−−−−√
[Pcosθ+Q]2+[Psinθ]2=(2n+1)2(P2+Q2)
P2cos2θ+2PQcosθ+Q2+P2sin2θ=(2n+1)2(P2+Q2)
P2+2PQcosθ+Q2=(2n+1)2(P2+Q2)
cosθ=(2n+1)2(P2+Q2)−P2−Q22PQ
cosθ=(4n2+4n+1)(P2+Q2)−P2−Q22PQ
cosθ=(4n2+4n+1)(P2+Q2)−1(P2+Q2)2PQ
cosθ=(4n2+4n+1–1)(P2+Q2)2PQ
cosθ=4n(n+1)(P2+Q2)2PQ(1)
Now if the angle between them is 90∘−θ , using the same configuration, when resolved, P⃗ and Q⃗ become,
P⃗ =Pcos(90−θ)i⃗ +Psin(90−θ)j⃗
Q⃗ =Qi⃗
From trigonometry, sin(90−θ)=cosθ and cos(90−θ)=sinθ . So,
P⃗ =Psinθi⃗ +Pcosθj⃗
Q⃗ =Qi⃗
Let's call the resultant of these vectors R90−θ→ .
R90−θ→=P⃗ +Q⃗ =[Psinθ+Q]i⃗ +Pcosθj⃗
∥∥∥R90−θ→∥∥∥=[Psinθ+Q]2+[Pcosθ]2−−−−−−−−−−−−−−−−−−−−√
From your question, ∥∥∥R90−θ→∥∥∥ is,
∥∥∥R90−θ→∥∥∥=(2n−1)P2+Q2−−−−−−−√
Equating the ∥∥∥R90−θ→∥∥∥ 's,
[Psinθ+Q]2+[Pcosθ]2−−−−−−−−−−−−−−−−−−−−√=(2n−1)P2+Q2−−−−−−−√
[Psinθ+Q]2+[Pcosθ]2=(2n−1)2(P2+Q2)
P2sin2θ+2PQsinθ+Q2+Pcos2θ=(2n−1)2(P2+Q2)
P2+2PQsinθ+Q2=(2n−1)2(P2+Q2)
sinθ=(2n−1)2(P2+Q2)−P2−Q22PQ
sinθ=(4n2–4n+1)(P2+Q2)−(P2+Q2)2PQ
sinθ=(4n2–4n+1–1)(P2+Q2)2PQ
sinθ=4n(n−1)(P2+Q2)2PQ(2)
Dividing (2) by (1)
sinθcosθ=4n(n−1)(P2+Q2)2PQ4n(n+1)(P2+Q2)2PQ
tanθ=4n(n−1)(P2+Q2)4n(n+1)(P2+Q2)
tanθ=n−1n+1
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Answer:
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