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weeeeeb [17]
2 years ago
14

A box sits at rest on a rough 33° inclined plane. Draw the free-body diagram, showing all the forces

Physics
1 answer:
Cloud [144]2 years ago
8 0

(a) The free body of all the forces include, frictional force, weight of the box acting perpendicular and another acting parallel to the plane.

(b) When the box is sliding down, the frictional force acts towards the right.

(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.

<h3>Free body diagram</h3>

The free body diagram of all the forces on the box is obtained by noting the upward force and downward forces on the box as shown below;

                      /  W2

                    Ф → Ff

                    ↓W1

where;

  • Ff is the frictional force resisting the down motion of the box
  • W1 is the perpendicular component of the box weight = Wcos(33)
  • W2 is the parallel component of the box weight = Wsin(33)

(b) When the box is sliding down, the frictional force acts towards the right.

(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.

Learn more about free body diagram of inclined objects here: brainly.com/question/4176810

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Mia is correct, but did not show a lot of work.

Explanation:

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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. T
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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 = \frac{y_w}{2.35}

         y_w = 2.35 cos 15

Post Weight

        sin 15 = \frac{x_p}{2.00}

         xp = 2.0 sin 15

Base weight

         cos (90-15) = \frac{x_b}{0.25}

         xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

     

          F_w \ y_w  + W_p \ x_p - W_b \ x_b = 0

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

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

     F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}

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

       

Let's  calculate

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

     F_w = 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,

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

Learn more  here: brainly.com/question/7031958

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Answer:

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Explanation:

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