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Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey witha force of 97.3 N, Jill pulls with 60.9 N in a direction 45° to the left, and Jane pulls in a direction 45 to the right with 159 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubbon?) Find the magnitude of the net force the people exert on the donkey. Number What is the direction of the net force? Express this as the angle from straight ahead between 0° and 90° with a positive sign for angles to the left and a negative sign for angles to the right.

Respuesta :

Answer:

most stubborn is Jill, F = 227.97 N and  θ = 43º  

Explanation:

As the forces give us we must use Newton's second law, in this case as the donkey does not move the acceleration is zero

Let's start by breaking down the forces

Jack

      F1 = 97.3 i ^ N

Jill

      F2 = 60.9 N

      θ = 45º      to the left

This angle measured from the positive part of the x axis is T ’= 90 + 45 = 135º

     F2x = F2 cos (θ’)

     F2y = F2 sin (θ’)

     F2x = 60.9 cos (135)

     F2y = 60.9 sin (135)

     F2x = -43.06 N

     F2y = 43.06 N

Jane

      F3 = 159N

      T = 45º

      F3x = F3 cos θ

     F3y = F3 sin θ

     F3x = 159 cos 45

     F3y = 159 sin 45

     F3x = 112.43 N

     F3y = 112.43 N

a) The most stubborn of the three must be the one who pulls in the opposite direction, bone Jill who is pulling to the left while the other two do it to the right

b) We perform the summary of the forces in each axis

X axis

     Fx = F1 + F2x + F3x

     Fx = 97.3 -43.06 +112.43

     Fx = 166.67 N

Axis y

     Fy = F2y + F3y

     Fy = 43.06 + 112.43

     Fy = 155.49 N

     F = √ Fx²+Fy²

     F = 166.67² + 155.49²

     F = 227.97 N

We use trigonometry

    tan θ = Fy / Fx

    θ = tan⁻¹ (155.49 / 166.67)

    θ = 43º

Regarding the x axis

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