Question

Two forces F1=5i+10j-20k and F2=10i-5j-15k act on a single point. The angle between F1 and F2 is nearly: a. 30° b. 45° c. 60° d. 90°

Answer 1

Answer:

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Answer 2

Answer:

The angle formed by the two forces is nearly equal to 45°.

Explanation:

Angle between two forces:

Given two forces are F₁=5i+10j-20k and F₂=10i-5j-15k

Angle between the two forces is given by

        cosθ= (F₁.F₂)/|F₁||F₂|

   where F₁.F₂ is the dot product of the two forces F₁ and F₂.

         F₁.F₂ = (5i+10j-20k).(10i-5j-15k)

                 = 50-50+300

                 = 300

            |F₁| = $\sqrt{5^{2}+10^{2}+(-20)^{2} }$

                  = $\sqrt{25+100+400}$

                  = $\sqrt{525}$

          |F₂| = $\sqrt{10^{2}+(-5)^{2}+(-15)^{2} }$

                = $\sqrt{100+25+225}$

                = $\sqrt{350}$

       Now substitute all the values in the formula

           cosθ = (F₁.F₂)/|F₁||F₂|

                    = 300/($\sqrt{525}\sqrt{350}$)

                    = 300/(5$\sqrt{21}$5$\sqrt{14}$)

                    = 12/$\sqrt{(21)(14)}$

                    = 12/7$\sqrt{6}$

           cosθ = 0.699  

                 θ = cos⁻¹(0.699)

                 θ ≅ cos⁻¹(cos45°)

                θ ≅ 45°

Hence the angle formed by the two forces is nearly equal to 45°.

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