Question

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

Answer 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

${R} = \sqrt{P^2 +Q^2 +2PQ cos\theta}$

SOLUTION:-

Here tangent means ${tan\theta}$

Since,

${tan\theta} = \dfrac{12}{5}$ 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

${R} = \sqrt{P^2 +Q^2 +2PQ cos\theta}$

${R} = \sqrt{ (130)^2 +(110)^2 + 2\times 130\times 110 \dfrac{5}{13}}$

${R} = \sqrt{ 16,900 +12,100 + 28,600 \dfrac{5}{13}}$

${R} = \sqrt{ 16,900 +12,100 + 2,200\times5}$

${R} = \sqrt{ 16,900 +12,100 + 11,000}$

${R} = \sqrt{40,000}$

${R} = \sqrt{(200)^2}$

R = 200N

So, resultant force acting on a body is 200N