To determine if the box will remain at rest or slide, we need to analyze the forces acting on it. Here’s the step-by-step solution:

Given Data:

• Mass of the box, $m = 20 \, \text{kg}$
• Gravitational acceleration, $g = 9.8 \, \text{m/s}^2$
• Static friction coefficient, $\mu_s = 0.6$
• Kinetic friction coefficient, $\mu_k = 0.5$
• Applied force, $F = 100 \, \text{N}$ at an angle of $30^\circ$

Step 1: Calculate the Normal Force ($N$)

The weight of the box is given by: $W = mg = 20 \times 9.8 = 196 \, \text{N}$

The applied force $F$ has a vertical component that affects the normal force: $F_y = F \sin(30^\circ) = 100 \times 0.5 = 50 \, \text{N}$

Since the force is angled upwards, it reduces the normal force. Thus: $N = W - F_y = 196 - 50 = 146 \, \text{N}$

Step 2: Calculate the Maximum Static Friction Force

The maximum static friction force $f_s$ that can resist the motion is: $f_s = \mu_s \cdot N = 0.6 \times 146 = 87.6 \, \text{N}$

Step 3: Determine the Horizontal Component of the Applied Force

The horizontal component of the applied force $F_x$ is: $F_x = F \cos(30^\circ) = 100 \times \frac{\sqrt{3}}{2} \approx 86.6 \, \text{N}$

Step 4: Compare $F_x$ with $f_s$

To determine if the box will slide, we compare $F_x$ with $f_s$:

• $F_x = 86.6 \, \text{N}$
• $f_s = 87.6 \, \text{N}$

Since $F_x < f_s$, the static friction is sufficient to keep the box at rest.

Conclusion

The box will remain at rest because the applied horizontal force is less than the maximum static friction force.

Do you want more details or have any questions?

Related Questions:

1. How would the result change if the angle of the applied force were increased?
2. What would happen if the mass of the box were increased to 30 kg?
3. How does the angle of the applied force affect the normal force?
4. What is the role of kinetic friction in this scenario?
5. How would this analysis change if the surface were inclined instead of horizontal?

Tip:

When analyzing friction problems, always separate the components of forces and compare the effective horizontal force to the frictional forces accurately.