The resultant of two forces 3p amd 2p is R. If the first force is doubled keeping same direction, then the resultant is also doubled. Find the angle between the two forces.
Answers
ANSWER:
angle between the two forces are120°
Explanation:
given that,
the resultant if two forces 3p and 2p is R
let the angle between them be θ
and
the forces be F1 and F2
and we know that,
resultant of two forces F1 and F2 inclined at the angle θ is given by,
R = √(F1² + F2² + 2F1F2cosθ) ....(1)
Here,
F1 = 3p
F2 = 2p
putting the values,
R = √{(3p) ² + (2p) ² + 2(3p)(2p) cosθ)}
R = √(9p² + 4p² + 12p² cosθ)
R² = 13p² + 12p²cosθ ....(2)
now,
also given that,
if the first force is doubled i.e.
6p
keeping the same direction, then the resultant is also doubled
so,
ACCORDING TO THE QUESTION
(2R)² = (6p)² + (2p)² + 2(6p)(2p) cosθ
4R² = 36p² + 4p² + 24p²cosθ
4R² = 40p² + 24p²cosθ
4R² = 4(10p² + 6p²cosθ)
R² = 10p² + 6p²cosθ ....(3)
now,
from (2) and (3)
we have,
13p² + 12p²cosθ = 10p² + 6p²cosθ
12p²cosθ - 6p²cosθ = 10p² - 13p²
6p²cosθ = -3p²
cosθ = -3p²/6p²
cosθ = -1/2
so,
θ = 120°
so,
angle between the two forces are
120°
Answer:
Angle between the two forces are 120°
Explanation:
Given Problem:
The resultant of two forces 3p amd 2p is R. If the first force is doubled keeping same direction, then the resultant is also doubled. Find the angle between the two forces.
Solution:
To Find:
The angle between the two forces.
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Method:
Resultant, R =√(a^2+b^2+2ab cosx)
By triangle law of vector addition,
So,
⇒r = √{(3p)^2+(2p)^2+2×3p×2p×cosx}
⇒Or, r = √(9p^2+4p^2+12p^2cosx)
⇒Or, r = √(13p^2+12p^2cosx)
According to the given problem:
⇒2r = √{(6p)^2+(2p)^2+2×6p×2p×cosx}
⇒2r =√(36p^2+4p^2+24p^2cosx)
⇒2r = √(40p^2+24p^2cosx)
⇒ 2{√(13p^2+12p^2cosx)}
⇒√(40p^2+24p^2cosx)
⇒4 (13p^2+12p^2cosx) = (40p^2+24p^2cosx)
⇒52p^2 + 48p^2cosx= 40p^2 + 24p^2cosx
⇒12 p^2 = - 24 p^2cosx
⇒cosx = - 1/2
⇒ x = cos^-1(-1/2)
⇒x = 120°
⇒So, angle between the forces is 120°.
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