A crate, in the form of a cube with edge lengths of $\mathbf{1}\mathbf{.2}\mathbf{\text{m}}$ , contains a piece of machinery; the center of mass of the crate and its contents is located $\mathbf{\text{0.30m}}$above the crate’s geometrical center. The crate rests on a ramp that makes an angle $\mathbf{\text{θ}}$with the horizontal. As$\mathbf{\text{θ}}$ is increased from zero, an angle will be reached at which the crate will either tip over or start to slide down the ramp. If the coefficient of static friction${\mathbf{\mu }}_{\mathbf{s}}$ between ramp and crate is $\mathbf{0}\mathbf{.}\mathbf{60}$(a) Does the crate tip or slide? And (b) at what angle $\mathbf{\theta }$does the crate tip or slide occur? (c) If${\mathbf{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0.70}$,does the crate tip or slide? And (d) If${\mathbf{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.70}$,at what angle u does the crate tip or slide occur?

1. The edge length of the cubeis $1.2\mathrm{m}$
2. The coefficient of static friction between ramp and crate is $0.60$, 0.70

Newton’s second law states that the force acting on the object is directly proportional to the acceleration of the object. The net force acting on the object is equal to the vector sum of all the forces acting on the object.

Using the formula for Newton’s second law, find if the crate tips or slides, with the givencoefficient of static friction between ramp and crate, is ${\mathbf{\mu }}_{\mathbf{S}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{60}\mathbf{.}$and${\mathbf{\mu }}_{\mathbf{S}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{70}\mathbf{.}$

Consider the equation for the angle.

$\mathrm{\theta }={\text{tan}}^{\text{-1}}{\mathrm{\mu }}_{\mathrm{s}}$

Let’s consider $\mathrm{W}$be the force of gravity on the crate. ${\mathrm{F}}_{\mathrm{N}}$is the normal force of the plane on the crate and $\mathrm{F}$ is the force of friction. Let’s consider $\mathrm{Y}$ axis is in the direction of normal force and $\mathrm{X}$axis is down the plane.

We have the x component according to Newton’s second law,

$\mathrm{W}\text{sin}\mathrm{\theta }-\mathrm{f}=0$ (i)

Similarly, we have the y component according to Newton’s second law,

${\mathrm{F}}_{\mathrm{N}}-\mathrm{Wcos\theta }=0$

${\mathrm{F}}_{\mathrm{N}}=\mathrm{Wcos\theta }$ (ii)

As the crate is about to slide,

$\mathrm{f}={\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}}={\mathrm{\mu }}_{\mathrm{S}}\mathrm{Wcos\theta }$

So, x component becomes,

$\begin{array}{c}\mathrm{W}\text{sin}\mathrm{\theta }-{\mathrm{\mu }}_{\mathrm{S}}\mathrm{W}\text{cos}\mathrm{\theta }=0\\ \text{tan}\mathrm{\theta }={\mathrm{\mu }}_{\mathrm{S}}\\ \mathrm{\theta }={\tan }^{-1}\left({\mathrm{\mu }}_{\mathrm{S}}\right)\\ ={\tan }^{-1}\left(0.60\right)\\ =31°\end{array}$

The torque associated with the force of friction tends to turn the crate clockwise and its magnitude is$\mathrm{fh}$, h is perpendicular distance from the bottom of the crate to the center of gravity of the crate. The torque associated with the normal force tends to turn the crate anticlockwise and its magnitude is$\frac{{\mathrm{F}}_{\mathrm{N}}\mathrm{l}}{2}$

As${\mathrm{F}}_{\mathrm{N}}=\mathrm{Wcos\theta }$and$\mathrm{f}=\mathrm{Wsin\theta }$

Therefore, we have,

$\mathrm{hf}=\frac{{\mathrm{F}}_{\mathrm{n}}\mathrm{l}}{2}$

$\mathrm{hWsin\theta }=\frac{\mathrm{lWcos\theta }}{2}$

This equation gives,

$\mathrm{\theta }={\tan }^{-1}\frac{\mathrm{l}}{2\mathrm{h}}$ (iii)

Edge of the cube is 1.2 m and center of gravity is located 0.3 m above the geometrical center of the cube. Therefore,

$\begin{array}{c}\mathrm{h}=\frac{1.2}{2}+0.3\\ =0.9\end{array}$

Substitute the value in equation (iii) to calculate $\mathrm{\theta }$

$\begin{array}{c}\mathrm{\theta }={\tan }^{-1}\frac{1.2\text{\hspace{0.17em}\hspace{0.17em}m}}{2\left(0.90\text{\hspace{0.17em}m}\right)}\\ =33.7°\end{array}$

Hence, as $\mathrm{\theta }$ increases from zero, the crate slides before it tips.

Therefore, the crate starts to slide when$\mathrm{\theta }=31°$As the crate is about to slide,

$\begin{array}{c}\mathrm{f}={\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}}\\ ={\mathrm{\mu }}_{\mathrm{S}}\mathrm{W}\text{cos}\mathrm{\theta }\end{array}$

So, x component becomes,

$\begin{array}{c}\mathrm{W}\text{sin}\mathrm{\theta }-{\mathrm{\mu }}_{\mathrm{S}}\mathrm{W}\text{cos}\mathrm{\theta }=0\\ \text{tan}\mathrm{\theta }={\mathrm{\mu }}_{\mathrm{S}}\\ \mathrm{\theta }={\tan }^{-1}\left({\mathrm{\mu }}_{\mathrm{S}}\right)\\ ={\tan }^{-1}\left(0.60\right)\\ =31°\end{array}$

For${\mathrm{\mu }}_{\mathrm{S}}=70$, we do the same calculations as above,

The crate begins to slide at,

$\begin{array}{c}\mathrm{\theta }={\tan }^{-1}\left({\mathrm{\mu }}_{\mathrm{s}}\right)\\ ={\tan }^{-1}\left(0.70\right)\\ =35°\end{array}$

When the crate is about to tip, there is no acceleration due to gravity. Therefore, we can write,

$\mathrm{f}=\mathrm{Wsin\theta }$ (iv)

Also,

${\mathrm{F}}_{\mathrm{N}}=\mathrm{Wcos\theta }$ (v)

Taking the ratio of these two equations, we get

role="math" localid="1663337768372" $\frac{\mathrm{f}}{{\mathrm{F}}_{\mathrm{N}}}=\mathrm{tan\theta }$ (vi)

The torque created due to force of friction is in the clockwise direction.

${\mathrm{\tau }}_{\mathrm{f}}=\mathrm{fh}$

$\mathrm{h}$ is the perpendicular distance from the bottom of the crate to the center of gravity.

The torque due to normal force crate the anticlockwise torque.

${\mathrm{\tau }}_{{\mathrm{F}}_{\mathrm{N}}}={\mathrm{F}}_{\mathrm{N}}\frac{\mathrm{l}}{2}$

For the equilibrium, both these torques must be same. Therefore,

$\begin{array}{c}\mathrm{fh}={\mathrm{F}}_{\mathrm{N}}\frac{\mathrm{l}}{2}\\ \frac{\mathrm{f}}{{\mathrm{F}}_{\mathrm{N}}}=\frac{\mathrm{l}}{2\mathrm{h}}\end{array}$

Comparing this equation with equation (vi), we get

$\begin{array}{c}\frac{\mathrm{l}}{2\mathrm{h}}=\mathrm{tan\theta }\\ \mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{l}}{2\mathrm{h}}\right)\\ =33.7°\end{array}$

Therefore, tipping begins at $33.7°$.