Physics Study Notes: Static Friction and Inclined Planes

Question 2: Static Friction and Inclined Planes

2.1 Definition of Static Frictional Force

• Static frictional force: The force that resists the initiation of sliding motion between two surfaces in contact. It acts in the opposite direction of the applied force until a maximum value is reached, beyond which motion begins. The static frictional force can be mathematically defined as:

  • $f<em>s \leq \mu</em>s N$
    where:

  • $f_s$ is the static frictional force,

  • $\mu_s$ is the coefficient of static friction,

  • $N$ is the normal force.

2.2 Free Body Diagram of the Crate

• Forces acting on the crate positioned on an inclined plane at 23.2°:

  • Weight ($W$) of the crate acting downwards:

  • $W = mg$

  • where:

    • $m = 95$ kg (mass of the crate)

    • $g = 9.81 \text{ m/s}^2$ (acceleration due to gravity)

  • Normal force ($N$) acting perpendicular to the surface of the plank.

  • Static frictional force ($f_s$) acting parallel to the surface, opposing the motion.

2.3 Calculation of the Magnitude of the Static Frictional Force

• To calculate the static frictional force ($f_s$) at 23.2°, we first resolve the weight of the crate into components:

  • Weight parallel to the incline: $W_{parallel} = W \sin(\theta) = mg \sin(23.2°)$

  • Weight perpendicular to the incline: $W_{perpendicular} = W \cos(\theta) = mg \cos(23.2°)$

  1. Calculation of Weight:

  • $W = 95 \text{ kg} \times 9.81 \text{ m/s}^2 = 931.95 ext{ N}$

  1. Calculation of Normal Force:

  • $N = W_{perpendicular} = 931.95 ext{ N} \cos(23.2°)$

  • Calculate: $N \approx 931.95 \text{ N} \times 0.9205 \approx 857.07 ext{ N}$

  1. Hence, the static frictional force is given by:

  • $f<em>s = \mu</em>s N$

  • Assuming maximum static friction where $f_s$ opposes the weight parallel:

  • $f_s = mg \sin(23.2°) \approx (95 \text{ kg}) \cdot (9.81 \text{ m/s}^2) \cdot 0.3940 \approx 373.35 ext{ N}$

2.4 Calculation of the Coefficient of Static Friction

• Coefficient of static friction ($\mu_s$) is calculated by:

  • $\mu<em>s = \frac{f</em>s}{N} = \frac{373.35 ext{ N}}{857.07 ext{ N}} \approx 0.436$

2.5 Change in Static Frictional Force with Angle Reduction

• Scenario: Plank is lowered to 20.0°

• The static frictional force is affected by the reduction in the angle of inclination.

• Since the component of weight acting parallel to the incline decreases, the static frictional force at the new angle will be:

  • Answer: LESS THAN the static frictional force at 23.2°.

2.6 Calculation of the Acceleration of the Crate with Kinetic Friction

• With oil on the surface, the crate begins to slide. The coefficient of kinetic friction ($\mu_k$) is given as half of the coefficient of static friction:

  • $\mu<em>k = \frac{1}{2} \mu</em>s = \frac{1}{2} \times 0.436 \approx 0.218$

• Forces on the crate include:

  • New static friction force (now kinetic) is:

  • The net force ( ext{F}_{net}) acting on the crate can be calculated as:

  • $F<em>{net} = W</em>{parallel} - f<em>k = mg \sin(\theta) - \mu</em>k N$

  • Substituting values for kinetic friction:

    • $= mg \sin(20.0°) - (\mu_k N)$

  • Calculate the net force on the crate and subsequently its acceleration $a$:

  • $F_{net} = 931.95 \text{ N} \sin(20.0°) - (0.218)(857.07 ext{ N})$

  • Calculate:

    • $F_{net} \approx 931.95 \text{ N} \cdot 0.3420 - 186.39 ext{ N} \approx 318.47 ext{ N}$

  • Now use Newton's second law to find acceleration:

    • $F<em>{net} = ma \Rightarrow a = \frac{F</em>{net}}{m} = \frac{318.47}{95} \approx 3.35 ext{ m/s}^2$

Conclusion

• The crate's static frictional force at 23.2° is calculated, the impact of adjusting the incline is analyzed, and the dynamics of sliding crates on various frictional surfaces are explored.