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spin [16.1K]
2 years ago
6

Determine the magnitude of the resultant force and its direction using both the parallelogram and Cartesian vector notation meth

ods. The direction of the resultant force is measured counter-clockwise from the positive x-axis. Draw the resultant force in the Cartesian coordinate system.F1= 600 N, F2= 900 N, β1 = 50 degree, and β2 = 40 degree.
Physics
1 answer:
Alika [10]2 years ago
8 0

Answer:

   F = 1494.52 N,   θ = 44º

Explanation:

For the sum of vectors by the parallelogram method, see attached, the vectors are drawn, the parallelogram is completed and a vector is drawn from the origin of the two vectors to the end point of the rectangle, this is the resulting vector.

The attachment shows this roughly.

For the Cartesian coordinate method, each vector is decomposed into its components, they are added algebraically and then the resulting vector is composed in the form of a module and angles

we use trigonometry to decompose the vectors.

The coordinate system can be seen in the attachment

           sin θ = y / R

           cos θ = x / R

            y = R sin θ

            x = R cos θ

Vector 1

module F₁ and angle β₁ = 50

            sin 50 = \frac{F_{1y} }{F_1}

            cos 50 = \frac{F_{1x} }{F_1}

            F_{1y} = F₁ sin 50

            F₁ₓ = F₁ cos 50

            F_{1y} = 600 sin 50 = 459.63 N

            F₁ₓ = 600 cos 50 = 385.67 N

Vector 2

modulus F₂ = 900N, angle β₂ = 40

            F_{2y} = 900 sin 40 = 578.51 N

            F₂ₓ = 900 cos 40 = 689.44 N

we find the resultant of each component

           F_{y} =F_{1y} + F_{2y}

           F_{y}  = 459.63 + 578.51

           F_{y}  = 1038.14 N

 

            Fₓ = F₁ₓ + F₂ₓ

            Fₓ = 385.67 + 689.44

             Fₓ = 1075.11 N

We use the Pythagorean theorem to find the modulus of the resultant

            F = Fₓ² + F_{y}^2

            F = √(1075.11² + 1038.14²)

            F = 1494.52 N

we use trigonometry for the angle

            tan θ = F_y / Fₓ

            θ = tan⁻¹ (F_y / Fₓ)

            θ = tan⁻¹ (1038.14 / 1075.11)

            θ = 44º

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Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

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          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

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Answer:

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If there is no energy loss due to air resistance, potential energy = kinetic energy

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Kinetic energy = 27 Joules

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Potential energy = 1.5 * 9.8 * 4

Potential energy = 58.8 Joules

From equation (1)

27 + Energy due to air resistance = 58.8

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By definition,
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