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A portable basketball set has a base and a post arrangement. The post arrangement consists of a post, backboard, hoop and net. The base can be filled with water to increase stability. The maximum weight of the base is about 810 N and the weight of the post arrangement is 26.0 N. The basketball set may topple over when the wind blows due to the large area of the backboard.
a) Determine the minimum force of the wind, Fw that will cause the basketball set to be blown over when it is at the angle shown. Ignore the effect of the wind on the base.
b) The base is filled with sand instead of water. The density of the sand is greater than the density of the water. Justify what would happen to the value of Fw calculated in part (a).

A Portable Basketball Set Has A Base And A Post Arrangement The Post Arrangement Consists Of A Post Backboard Hoop And Net The Base Can Be Filled With Water To class=

Sagot :

The rotational equilibrium condition allows finding the response to the minimum force of the wind and what happens when changing the water for sand, in the system

  a) The minimum force of the wind that turns the system is Fw = 17.64 N

  b) The system resists much greater forces because the base has more mass

 Newton's Second Law can be applied to rotational motion in this case when the angular acceleration is zero we have the special case of rotational equilibrium

               Σ τ = 0

Where τ is the torque  

The reference system is a coordinate system with respect to which the torques are measured, in this case we will fix the system at the turning point, the junction of the base and the pole, we will assume that the counterclockwise rotations are positive.

For the torque the distance used is the perpendicular distance from the direction of the force to the axis of rotation, let's find this distance for each force

Wind force

         cos 15 = [tex]\frac{y_w}{2.35}[/tex]

         [tex]y_w[/tex] = 2.35 cos 15

Post Weight

        sin 15 = [tex]\frac{x_p}{2.00}[/tex]

         xp = 2.0 sin 15

Base weight

         cos (90-15) = [tex]\frac{x_b}{0.25}[/tex]

         xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

     

          [tex]F_w \ y_w + W_p \ x_p - W_b \ x_b = 0[/tex]

a) To calculate the minimum wind force we substitute the given values

They indicate the weight of the post is [tex]W_p[/tex] = 26.0 N and the weight of the base with water is [tex]W_b[/tex] = 810 N

     [tex]F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}[/tex]

     [tex]F_w = \frac{W_b \ 0.25 cos75 \ - W_p \ 2 sin 15}{2.35 cos 15}[/tex]

       

Let's  calculate

     [tex]F_w = \frac{810 \ 0.25 \ cos75 \ - 26.0 \ 2 \ sin 15}{2.35 cos15}\\F_w = \frac{52.41 - 10.30}{2.3699}[/tex]

     [tex]F_w[/tex] = 17.64 N

b) The water is exchanged for sand.

In this case, as the density of the sand is greater than that of the water, the base will have more weight, so it will resist stronger winds before turning over.

Using the rotational equilibrium condition we can find the response to the minimum force of the wind and what happens when changing the water for sand,

  a) the minimum force of the wind that turns the system is Fw = 17.64 N

  b) the system resists much greater forces because the base has more mass

Learn more  here: brainly.com/question/7031958

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