To solve this problem, we need to analyze the forces acting on both blocks and apply the conditions for static equilibrium, since we are looking for the minimum mass $m$ that will prevent slipping.

Step 1: Analyzing the System

• The inclined block of mass $m$ rests on a 20° slope.
• The block is connected to a hanging mass $m_2 = 2.0 \ \text{kg}$.
• Coefficient of static friction between the inclined block and the surface: $\mu_s = 0.82$.
• Coefficient of kinetic friction (in case it starts moving): $\mu_k = 0.51$.
• To prevent slipping, the tension in the string should balance the static friction and the component of gravity parallel to the slope.

Step 2: Forces on the Inclined Block

For the inclined block:

• Weight: $mg$.
• Normal force: $N = mg \cos(20^\circ)$.
• Static friction force (maximum): $f_s = \mu_s N = \mu_s mg \cos(20^\circ)$.
• Gravitational force component parallel to the slope: $mg \sin(20^\circ)$.

Step 3: Forces on the Hanging Block

For the hanging block:

• Weight: $m_2 g = 2.0 \ \text{kg} \cdot g$.

The tension $T$ in the string is the same on both sides due to the massless pulley.

Step 4: Condition for Static Equilibrium

For the system to be in equilibrium (the inclined block should not slip):

• The tension $T$ should balance the friction force and the component of gravity pulling the inclined block down the slope: $T \leq f_s + mg \sin(20^\circ).$
• Simultaneously, $T$ must balance the weight of the hanging mass: $T = m_2 g.$

Combining the two equations: $m_2 g \leq \mu_s mg \cos(20^\circ) + mg \sin(20^\circ).$

Step 5: Solve for $m$

Let's isolate $m$: $m_2 \leq m \left(\mu_s \cos(20^\circ) + \sin(20^\circ)\right).$ $m \geq \frac{m_2}{\mu_s \cos(20^\circ) + \sin(20^\circ)}.$

Substitute the given values:

• $m_2 = 2.0 \ \text{kg}$
• $\mu_s = 0.82$
• $\cos(20^\circ) \approx 0.9397$
• $\sin(20^\circ) \approx 0.3420$

$m \geq \frac{2.0}{0.82 \cdot 0.9397 + 0.3420}.$

Step 6: Calculate the Minimum Mass

Let's evaluate this expression:

$m \geq \frac{2.0}{0.82 \cdot 0.9397 + 0.3420} \approx \frac{2.0}{1.1093} \approx 1.80 \ \text{kg}.$

Final Answer

The minimum mass $m$ that will stick and not slip on the slope is approximately:

$\boxed{1.80 \ \text{kg}}$

This means that any mass less than this would result in the block slipping down the slope due to insufficient friction.

Would you like more details on any part of this solution or have any questions?

Related Questions:

1. What would happen if the coefficient of kinetic friction $\mu_k$ was used instead of $\mu_s$?
2. How does the angle of the incline affect the minimum mass $m$?
3. What is the effect of increasing the mass of the hanging block to 3 kg?
4. How would the solution change if the pulley had mass or friction?
5. What is the tension in the string if $m$ is exactly 1.80 kg?

Tip:

When dealing with inclined planes and friction, always decompose the forces into components parallel and perpendicular to the slope for easier calculations.