An engineer is designing a conveyor system for loading hay bales into a wagon (Given figure). Each bale is$\mathbf{0}\mathbf{.25}\mathbf{\text{\hspace{0.17em}m}}$wide,$\mathbf{0.50}\mathbf{\text{\hspace{0.17em}m}}$high, and$\mathbf{0.80}\mathbf{\text{\hspace{0.17em}m}}$long (the dimension perpendicular to the plane of the figure), with mass $\mathbf{30.0}\mathbf{\text{\hspace{0.17em}kg}}$. The center of gravity of each bale is at its geometrical center. The coefficient of static friction between a bale and the conveyor belt is$\mathbf{0.60}$, and the belt moves with constant speed. (a) The angle $\mathbf{\beta }$of the conveyor is slowly increased. At some critical angle a bale will tip (if it doesn’t slip first), and at some different critical angle it will slip (if it doesn’t tip first). Find the two critical angles and determine which happens at the smaller angle. (b) Would the outcome of part (a) be different if the coefficient of friction were $\mathbf{0.40}$?

The force that surfaces apply to prevent solid objects from passing through one another is known as the normal force.

Here we have, hay bales into a wagon. Each bale is$0.25\text{\hspace{0.17em}m}$wide,$0.50\text{\hspace{0.17em}m}$high, and$0.8\text{\hspace{0.17em}m}$long.

Mass of bale is$30\mathrm{kg}$.

The coefficient of static friction between a bale and the conveyor belt is $0.6$

Also, the belt moves at a constant speed.

a.

Before $\beta$increases, the first angle is between the beam of the weight of the center of the bale and the normal force exerted by the wagon as shown in the above figure. So, we can find this angle by,

Now, when $\mathrm{\beta }$increases, the summation of vertical forces is zero, so we can get it by

$\mathrm{f}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{mgcos\beta }$ (1)

Now, for the horizontal force,

Put the value of${\mathrm{\mu }}_{\mathrm{s}}$in the above equation. We get,

Hence, the two critical angles are $\mathrm{\beta }=27°\text{and}31°$

(b)

By part (a), we have$\mathrm{\beta }={\tan }^{-1}{\mathrm{\mu }}_{\mathrm{s}}$

So, when the coefficient of friction is $0.40$, $\mathrm{\beta }$ is given by

Hence, when the coefficient of friction is$0.40$ ,$\mathrm{\beta }=22°$