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 =
= 2.35 cos 15
Post Weight
sin 15 =
xp = 2.0 sin 15
Base weight
cos (90-15) =
xB = 0.25 cos 75
Let's substitute in the rotational equilibrium equation
a) To calculate the minimum wind force we substitute the given values
They indicate the weight of the post is = 26.0 N and the weight of the base with water is
= 810 N
Let's calculate
= 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
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