## Task 30/63

On an inclined plane of adjustable inclination $\alpha$, a mass $m$ is placed. There is static friction between the mass and the plane. The coefficient of static friction is $\mu_{0}$. The mass is attached to the upper edge of the inclined plane by a thread. This thread breaks under a load $L_{r}$. The inclination of the plane is now increased (starting from $0^{\circ}$).

At what inclination $\alpha_{r}$ will the thread break?

a) Solve the problem in general terms, i.e., explicitly express $\alpha_{r}=f\left(m ; \mu_{0} ; L_{r} ; g\right)$.

b) Calculate $\alpha_{r}$ numerically for $m=3 \mathrm{~kg}, \mu_{0}=0.3, L_{r}=8 \mathrm{~N}$, and $g=9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}$.
Driving force: $P_{A}=m g \sin \alpha$

Friction force: $P_{R}=\mu_{0} m g \cos \alpha$

Thread load: $L=m g\left(\sin \alpha-\mu_{0} \cos \alpha\right)$

Rupture condition: $L_{r} \geq m g\left(\sin \alpha-\mu_{0} \cos \alpha_{r}\right)$

The rupture condition results in an equation of the general form $\sin x-a \cos x=b$ with $x=\alpha_{r}, a=\mu_{0}$ and $b=\frac{L_{r}}{m g}$.

If one sets $a=\tan y$ and multiplies the equation by $\cos y$, the following is obtained and further with an addition theorem

$$
\sin x \cos y-\sin y \cos x=b \cos y \rightarrow \sin (x-y)=b \cos y
$$

Solving for $x$, one gets $x=y+\arcsin b \cos y$. Substituting $y$ back, one obtains

$$
x=\arctan a+\arcsin \left(\frac{b}{\sqrt{1+a^{2}}}\right)
$$

Thus, the solution and the desired numerical value are

$$
\alpha_{r}=\arctan \mu_{0}+\arcsin \left(\frac{L_{r}}{m g \sqrt{1+\mu_{0}^{2}}}\right) \approx 31^{\circ} 48^{\prime}
$$