Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,2.50kN of fo

Question

Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,2.50kN of force, the tide provides 1, point, 20, k, N,1.20kN of force from the direction 30, point, 0, degrees,30.0 ∘ more northerly than the wind. Give your answer to 2 significant figures. Remember that ‘an easterly wind’ means a wind coming from the East

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Thu Giang 4 years 2021-07-14T10:46:23+00:00 1 Answers 118 views 0

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  1. bonexptip
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    2021-07-14T10:47:47+00:00
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    Answer:

    F = 3.6 kN, direction is 9.6º to the North – East

    Explanation:

    The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

    Let’s use trigonometry to find the component of the forces, let’s use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

    Wind

    X axis

              F₁ = 2.50 kN

    Tide

             cos 30 = F₂ₓ / F₂

             sin 30 = F_{2y} / F₂

              F₂ₓ = F₂ cos 30

             F_{2y} = F₂ sin 30

             F₂ₓ = 1.20cos 30 = 1.039 kN

             F_{2y} = 1.20 sin 30 = 0.600 kN

    the resultant force is

    X axis

            Fₓ = F₁ₓ + F₂ₓ

            Fₓ = 2.50 +1.039

            Fₓ = 3,539 kN

            F_y = F_{2y}

            F_y = 0.600

    to find the vector we use the Pythagorean theorem

             F = \sqrt{F_x^2 +F_y^2}

             F = \sqrt{ 3.539^2 + 0.600^2 }

             F = 3,589 kN

    the address is

             tan θ = F_y / Fₓ

             θ = tan⁻¹ \frac{F_y}{F_x}

             θ = tan⁻¹  \frac{0.6}{3.539}0.6 / 3.539

             θ = 9.6º

    the resultant force to two significant figures is

             F = 3.6 kN

    the direction is 9.6º to the North – East

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