Understanding Forces on an Inclined Plane
When an object sits on an inclined plane, gravity acts downwards, but we decompose it into two components:
- Parallel component ($mg \sin \theta$): Pulls the object down the slope.
- Perpendicular component ($mg \cos \theta$): Presses the object into the surface, creating the normal force $N = mg \cos \theta$.
When we apply a force $F$ to push the block up the plane, friction acts down the plane to oppose the motion. At the threshold of moving upwards, the forces are in equilibrium.
The Physics Problem
Given:
- Mass $m = 10 \text{ kg}$
- Angle $\theta = 30^{\circ}$
- Upward Force $F = 100 \text{ N}$
- Acceleration due to gravity $g \approx 9.8 \text{ m/s}^2$ (or $10 \text{ m/s}^2$ for simplicity)
Step-by-Step Solution
Force Balance Equation: To move the block up, the applied force $F$ must overcome both the parallel component of gravity and the maximum static friction $f_s$. $$F = mg \sin \theta + f_s$$
Express Friction: The maximum static friction is $f_s = \mu N$, where $N = mg \cos \theta$. $$F = mg \sin \theta + \mu mg \cos \theta$$
Plug in the values: Using $g = 9.8 \text{ m/s}^2$: $$100 = (10 \times 9.8 \times \sin 30^{\circ}) + \mu (10 \times 9.8 \times \cos 30^{\circ})$$ $$100 = (98 \times 0.5) + \mu (98 \times 0.866)$$ $$100 = 49 + 84.87\mu$$
Solve for $\mu$: $$51 = 84.87\mu$$ $$\mu = \frac{51}{84.87} \approx 0.60$$
Conclusion
The coefficient of friction $\mu$ is approximately 0.60.