A flag pole consists of a 2.0 m long, 5.0 kg decorative ball on the end. What is the gravitational torque on the flag pole, calculated about an axis at the fixed end of the pole?
Thanks for the quick help
To calculate the gravitational torque on the flag pole, we need to consider the following equation:
τ = r * F * sin(θ)
Where:
τ is the torque
r is the perpendicular distance from the axis of rotation to the line of action of the force
F is the force applied
θ is the angle between the force vector and the line connecting the axis to the point of application of force
In this case, the force is the weight of the decorative ball, which is equal to the mass of the ball multiplied by the acceleration due to gravity:
F = m * g
Where:
m is the mass of the decorative ball
g is the acceleration due to gravity (approximately equal to 9.8 m/s^2 on Earth)
In addition, the perpendicular distance from the axis of rotation to the line of action of the force is the length of the flag pole, which is given as 2.0 m.
By substituting the values into the equation, we can calculate the gravitational torque:
τ = r * F * sin(θ)
= 2.0 m * (5.0 kg * 9.8 m/s^2) * sin(θ)
= 98 N * sin(θ)
Note that the angle θ is not mentioned in the question, so we are unable to determine the precise value of the torque without that information.