Question

You have a flag pole at an angle of 45 degrees to a wall it is designed to support a max torque of 40Nm. You bought a new one with a mass of 8.75kg and the center of mass is 0.750m from the bracket will it support it and at what angle could the new flag pole be placed at to reduce the torque so it can be used?

Answer 1

Answer:

0.915 rad or 52.44 degrees

Explanation:

Let g = 10 m/s2. We can calculate the gravity force acting on center of mass of the new 8.75 kg pole:

$F = mg = 8.75 * 10 = 87.5 kg$

The moment is the dot product of force and moment arm

$M = \vec{F} \cdot \vec{r} = Frcos\theta = 87.5*0.75cos\theta = 65.625cos\theta$

where θ is the angle between the gravity force and the pole, or between the pole and the wall. As M can be at its max, which is 40 Nm, we can solve for θ:

$62.625cos\theta = 40$

$cos\theta = 40 / 62.625 = 0.61$

$\theta = cos^{-1}0.61 = 0.915 rad = 0.915 * 180 /\pi = 52.44^o$