Problem Statement
Analyze the circuit: Analyze the circuit: Three forces act on a particle in equilibrium: $F_1 = 10$ N (north), $F_2 = 10$ N (south), and $F_3 = ?$. Find $F_3$. Equilibrium: vector sum = 0 Step 1: $F_1 + F_2 = 10\hat{j} – 10\hat{j} = 0$. Step 2: For equilibrium, $F_3 = 0$ N (no third force is needed). $$\boxed{F_3 = 0}$$
Given Information
- $. Equilibrium: vector sum = 0 Step$
Physical Concepts & Formulas
Newton’s second law $\mathbf{F}_\text{net} = m\mathbf{a}$ is the fundamental relation between net force and acceleration. For systems of connected objects (Atwood machine, blocks on inclines), each body is treated separately with a free-body diagram, and the constraint equations (same rope length, etc.) link the accelerations.
- $\mathbf{F}_{\text{net}} = m\mathbf{a}$ — Newton’s second law
- Atwood: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$
- $f_k = \mu_k N$ — kinetic friction
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
$$\boxed{F_3 = 0}$$
$$\mathcal{E} – I(R_{eq} + r) = 0 \implies I = \frac{\mathcal{E}}{R_{eq}+r}$$
$$R_{\text{parallel}} = \frac{3\times6}{3+6} = 2\,\Omega$$
Answer
$$\boxed{F_3 = 0}$$
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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