Problem Statement
Solve the Newton’s Laws / mechanics problem: Solve the Newton’s Laws / mechanics problem: A water drop on a tilted surface will slide when the tilt angle $\alpha$ exceeds a critical value. Derive the condition in terms of advancing ($\theta_a$) and receding ($\theta_r$) contact angles. A drop slides when gravitational force exceeds the contact
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Friction is a contact force opposing relative motion (kinetic friction) or impending motion (static friction). On an inclined plane, the weight component along the slope is $mg\sin\theta$ and the normal force is $N = mg\cos\theta$, giving maximum static friction $f_{s,\max} = \mu_s mg\cos\theta$. The condition for sliding is $\tan\theta > \mu_s$.
- $f = \mu N$ — kinetic friction force
- $N = mg\cos\theta$ — normal force on incline
- $mg\sin\theta – \mu mg\cos\theta = ma$ — Newton’s 2nd law along incline
- $\tan\theta_c = \mu_s$ — critical angle for sliding
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\sum F_x = ma_x\quad,\quad \sum F_y = ma_y = 0\quad\text{(if no vertical acceleration)}$$
$$a = \frac{(m_2-m_1)g}{m_1+m_2} = \frac{(5-3)\times9.8}{8} = \frac{19.6}{8} = 2.45\,\text{m/s}^2$$
$$T = \frac{2m_1 m_2 g}{m_1+m_2} = \frac{2\times3\times5\times9.8}{8} = \frac{294}{8} = 36.75\,\text{N}$$
Answer
$$\boxed{a = \dfrac{(m_2-m_1)g}{m_1+m_2}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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