Solution to Question 28
1. Force Analysis
Let the position of the disc be at an angle $\theta$ from the upward direction of the line of greatest slope. The forces acting on the disc in the plane of the incline are:
Let the position of the disc be at an angle $\theta$ from the upward direction of the line of greatest slope. The forces acting on the disc in the plane of the incline are:
- Gravity Component ($G$): $mg \sin \alpha$, acting vertically down the slope.
- Spring Force ($T$): $k \Delta r$. For equilibrium in this configuration (mass above nail), the spring is compressed and pushes the disc outward (direction $\theta$ from Up).
- Friction ($f$): Balances the resultant of the other two forces.
2. Vector Relationship
For equilibrium, friction must balance the vector sum of the spring force and gravity: $$ \vec{f} + \vec{T} + \vec{G} = 0 \implies |\vec{f}| = |\vec{T} + \vec{G}| $$ The condition for the disc not to slide is that the required friction must not exceed the limiting friction $f_{max}$: $$ |\vec{T} + \vec{G}| \le f_{max} $$ Given that $\mu \approx \tan \alpha$, we have $f_{max} = \mu mg \cos \alpha \approx mg \sin \alpha = G$. So the condition becomes: $$ |\vec{T} + \vec{G}| \le G $$
For equilibrium, friction must balance the vector sum of the spring force and gravity: $$ \vec{f} + \vec{T} + \vec{G} = 0 \implies |\vec{f}| = |\vec{T} + \vec{G}| $$ The condition for the disc not to slide is that the required friction must not exceed the limiting friction $f_{max}$: $$ |\vec{T} + \vec{G}| \le f_{max} $$ Given that $\mu \approx \tan \alpha$, we have $f_{max} = \mu mg \cos \alpha \approx mg \sin \alpha = G$. So the condition becomes: $$ |\vec{T} + \vec{G}| \le G $$
3. Resultant Calculation
We calculate the magnitude of the resultant vector $\vec{R} = \vec{T} + \vec{G}$.
We calculate the magnitude of the resultant vector $\vec{R} = \vec{T} + \vec{G}$.
- Vector $\vec{G}$ acts Down ($180^\circ$ from Up).
- Vector $\vec{T}$ acts at angle $\theta$ from Up.
4. Solving for Deformation
Applying the stability condition $R^2 \le G^2$: $$ T^2 + G^2 – 2TG \cos\theta \le G^2 $$ Subtracting $G^2$ from both sides: $$ T^2 – 2TG \cos\theta \le 0 $$ $$ T(T – 2G \cos\theta) \le 0 $$ Since $T > 0$, we must have: $$ T – 2G \cos\theta \le 0 \implies T \le 2G \cos\theta $$ Substituting $T = k \Delta r$ and $G = mg \sin\alpha$: $$ k \Delta r \le 2 mg \sin\alpha \cos\theta $$ Thus, the suitable expression for the deformation limit is: $$ \Delta r = \frac{2 mg \sin\alpha}{k} \cos\theta $$
Applying the stability condition $R^2 \le G^2$: $$ T^2 + G^2 – 2TG \cos\theta \le G^2 $$ Subtracting $G^2$ from both sides: $$ T^2 – 2TG \cos\theta \le 0 $$ $$ T(T – 2G \cos\theta) \le 0 $$ Since $T > 0$, we must have: $$ T – 2G \cos\theta \le 0 \implies T \le 2G \cos\theta $$ Substituting $T = k \Delta r$ and $G = mg \sin\alpha$: $$ k \Delta r \le 2 mg \sin\alpha \cos\theta $$ Thus, the suitable expression for the deformation limit is: $$ \Delta r = \frac{2 mg \sin\alpha}{k} \cos\theta $$