Problem Analysis:

We are tasked with finding the two possible values of the angle $\alpha$ for a box in limiting equilibrium on an inclined slope. Here's the given information:

1. Mass of the box ($m$): $2.5 \, \text{kg}$
2. Inclination angle ($\alpha$): To be determined
3. Coefficient of friction ($\mu$): $0.15$
4. Horizontal force ($F_h$): $10 \, \text{N}$
5. Limiting equilibrium: The box is at the verge of sliding, so frictional force is at its maximum, $F_r = \mu R$, where $R$ is the normal reaction.

The forces acting on the box are:

• Weight ($W = mg = 2.5 \cdot 9.8 = 24.5 \, \text{N}$) acting vertically downward.
• Horizontal force ($F_h = 10 \, \text{N}$).
• Friction ($F_r$), resisting motion.
• Normal reaction ($R$) perpendicular to the slope.

Using these, we'll resolve forces along and perpendicular to the slope, and apply the limiting equilibrium conditions.

Force Equations:

1. Perpendicular to the slope: $R = W \cos\alpha + F_h \sin\alpha$

2. Along the slope: In limiting equilibrium, friction balances the other forces: $F_r = W \sin\alpha - F_h \cos\alpha$ And since $F_r = \mu R$, substituting $R$: $\mu (W \cos\alpha + F_h \sin\alpha) = W \sin\alpha - F_h \cos\alpha$

Solve for $\alpha$:

Rearrange the equation to isolate $\tan\alpha$: $\mu (W \cos\alpha + F_h \sin\alpha) = W \sin\alpha - F_h \cos\alpha$ Expand and collect terms: $\mu W \cos\alpha + \mu F_h \sin\alpha = W \sin\alpha - F_h \cos\alpha$ $\mu W \cos\alpha + F_h \cos\alpha = W \sin\alpha - \mu F_h \sin\alpha$ Factorize terms involving $\sin\alpha$ and $\cos\alpha$: $(W + \mu F_h) \sin\alpha = (\mu W + F_h) \cos\alpha$ Divide through by $\cos\alpha$ (valid since $\cos\alpha \neq 0$): $\tan\alpha = \frac{\mu W + F_h}{W + \mu F_h}$

Substitute Values:

$W = 24.5 \, \text{N}, \, F_h = 10 \, \text{N}, \, \mu = 0.15$

$\tan\alpha = \frac{(0.15)(24.5) + 10}{24.5 + (0.15)(10)}$

Simplify: $\tan\alpha = \frac{3.675 + 10}{24.5 + 1.5} = \frac{13.675}{26}$

$\tan\alpha \approx 0.5267$

Find $\alpha$: $\alpha_1 = \arctan(0.5267) \approx 27.77^\circ$

Since there are two possible orientations for limiting equilibrium (sliding up or down), the second solution is: $\alpha_2 = 180^\circ - \alpha_1 = 180^\circ - 27.77^\circ = 152.23^\circ$

Final Answer:

The two possible values of $\alpha$ are approximately: $\boxed{27.8^\circ \, \text{and} \, 152.2^\circ.}$

Would you like a more detailed breakdown of the steps or clarification on any part of the solution?

Related Questions:

1. What happens to the angle $\alpha$ if the coefficient of friction is increased?
2. How does the horizontal force $F_h$ affect the limiting equilibrium condition?
3. What if the box's mass were doubled—how would $\alpha$ change?
4. Can we determine the minimum horizontal force $F_h$ required to move the box up the slope?
5. What role does the direction of the horizontal force play in determining the equilibrium angle?

Tip:

Always draw a clear free-body diagram to identify forces acting along and perpendicular to the slope.