Find the resultant of two forces 130 N and 110 N respectively, acting at an angle whose tangent is 12/5.​

1 thought on “Find the resultant of two forces 130 N and 110 N respectively, acting at an angle whose tangent is 12/5.​”

1. Topic :-

   Force

   Answer :-

   200N

   Given:-

   The two forces acting on a body is 130N and 110N and tangent angle is 12/5

   To find :-

   Resultant of the forces

   Formula to know :-

   For finding the resultant forces acting on a body is

   [tex]{R} = \sqrt{P^2 +Q^2 +2PQ cos\theta}[/tex]

   SOLUTION:-

   Here tangent means [tex]{tan\theta}[/tex]

   Since,

   [tex]{tan\theta} = \dfrac{12}{5}[/tex] But in formula we have costheta From, this we can find costheta

   tanA = opp/adj

   Since ,

   opposite side = 12

   Adjacent side = 5

   From Pythagoras theorem we can find hypotenuse

   (opp)² + (adj)² = (hyp)²

   (12)² + (5)² = (hyp)²

   144 + 25 = (hyp)²

   169 = (hyp)²

   hypotenuse = 13

   Now,

   All we know that

   cosA = adj/hyp So,

   cosA = 5/13

   Calculations:-

   Now , we can find resultant forces

   [tex]{R} = \sqrt{P^2 +Q^2 +2PQ cos\theta}[/tex]

   [tex]{R} = \sqrt{ (130)^2 +(110)^2 + 2\times 130\times 110 \dfrac{5}{13}}[/tex]

   [tex]{R} = \sqrt{ 16,900 +12,100 + 28,600 \dfrac{5}{13}}[/tex]

   [tex]{R} = \sqrt{ 16,900 +12,100 + 2,200\times5}[/tex]

   [tex]{R} = \sqrt{ 16,900 +12,100 + 11,000}[/tex]

   [tex]{R} = \sqrt{40,000}[/tex]

   [tex]{R} = \sqrt{(200)^2}[/tex]

   R = 200N

   So, resultant force acting on a body is 200N

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