FOCUSED REVIEW
Focused Review — Static Equilibrium and Stability — Calculus-Based
Reinforce the highest-leverage ideas and representative problem-solving tools for Static Equilibrium and Stability.
TIME
Approximately 15 minutes
BEST FOR
Targeted reinforcement
FINISH WITH
A readiness check
After this focused review, you'll be able to...
reinforce the key relationships, apply them to representative problems, and identify what still needs work.
Choose how you want to review
Course Alignment
This Physics Sensei Unit Review is an independent learning resource. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.
RESOURCE: Physics Sensei Unit Review | UNIT ID: MEC-U23 | TOPIC: Static Equilibrium and Stability | COURSE LEVEL: Calculus-Based
BEST USED ✓ After learning the unit ✓ Before starting homework ✓ Before a quiz or exam
Your Review Plan
Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.
6 Stages • Approximately 15 minutes.
Warm-Up Check
Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.
ACTIVITY 1
Vector Conditions
Answer before revealing the response.
State the equilibrium equations for a rigid body.
Reveal Answers
ΣF = 0 and Στ = 0, with τ = r × F.
Why it works: This checks a prerequisite idea used in the equilibrium and stability review.
ACTIVITY 2
Energy Criterion
Answer before revealing the response.
What condition identifies an equilibrium configuration in U(q)?
Reveal Answers
dU/dq = 0 for the generalized coordinate q.
Why it works: This checks a prerequisite idea used in the equilibrium and stability review.
ACTIVITY 3
Quick Application
Answer before revealing the response.
If d2U/dq2 > 0 at an equilibrium point, classify it.
Reveal Answers
Stable; the potential energy has a local minimum.
Why it works: This checks a prerequisite idea used in the equilibrium and stability review.
Ready to strengthen your understanding?
You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?
Core Concepts
Reinforce the two highest-leverage relationships, then use them in representative situations.
KEY CONCEPT 1
Vector Force and Torque Balance
Rigid-body equilibrium requires both translational and rotational equations. In three dimensions, the vector equations supply up to six scalar conditions. Torque about an origin is the sum of r × F for all external forces.
ΣF = 0 and Στ = 0, with τ = r × F
Example: A force whose line of action passes through the chosen origin contributes zero torque about that origin.
Sensei note: Changing the origin changes individual torques, but for an equilibrium system the total torque remains zero.
KEY CONCEPT 2
Potential-Energy Stability
For a conservative one-coordinate system, equilibrium occurs at dU/dq = 0. A positive second derivative indicates a local minimum and stable equilibrium; a negative second derivative indicates a local maximum and unstable equilibrium.
dU/dq = 0; stable if d2U/dq2 > 0
Example: Near a stable equilibrium q0, U can be approximated by a quadratic minimum.
Sensei note: If the second derivative is zero, higher-order terms may be needed to classify the equilibrium.
KEY CONCEPT 3
Small-Angle Stability and Restoring Torque
Near a stable equilibrium, expand U about q0. If the leading nonzero curvature is positive, the restoring generalized force is approximately proportional to −(q − q0). This connects static stability to small oscillations.
Q ≈ −k(q − q0) near a stable equilibrium
Example: With U ≈ U0 + 1/2k(q − q0)2, the generalized restoring force is approximately −k(q − q0).
Sensei note: A positive curvature gives a restoring response; a negative curvature drives the system away.
Ready to apply these ideas?
You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?
Guided Practice
Apply the reinforced ideas to two representative situations, then use the strategy card to check your setup.
PRACTICE 1
Cable Tension Derivation
Apply torque balance about the hinge.
A horizontal beam of length L and weight W is hinged at one end and supported by a cable with tension T at angle θ to the beam at the far end. Derive T.
Reveal Answers
Torque balance about the hinge gives TL sin θ − W(L/2) = 0, so T = W/(2 sin θ).
Why it works: The hinge contributes no torque about itself, so cable torque balances the beam-weight torque.
PRACTICE 2
Energy Stability
Differentiate U(q) and classify the equilibrium.
A system has U(q) = aq2 + bq4 with a > 0 and b > 0. Classify q = 0.
Reveal Answers
dU/dq = 2aq + 4bq3, so q = 0 is an equilibrium. d2U/dq2 = 2a + 12bq2, which is positive at q = 0, so the equilibrium is stable.
Why it works: The first derivative vanishes and the positive second derivative gives a local minimum.
PRACTICE 3
Angular Potential Energy
Use first and second derivatives.
For U(θ) = U0 + A(1 − cos θ), with A > 0, classify θ = 0 and θ = π.
Reveal Answers
dU/dθ = A sin θ, so both are equilibria. d2U/dθ2 = A cos θ: positive at 0, so stable; negative at π, so unstable.
Why it works: The curvature is positive at θ = 0 and negative at θ = π.
Ready to check your understanding?
You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?
Confidence Check
You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.
QUICK CHECK 1
Vector Torque
Evaluate the direction.
If r points in +x and F points in +y, what is the direction of τ = r × F?
Reveal Answers
+z by the right-hand rule.
Why it works: The cross product +x × +y points in +z.
QUICK CHECK 2
Stability from Curvature
Classify the point.
At q = q0, dU/dq = 0 and d2U/dq2 < 0. What type of equilibrium is present?
Reveal Answers
Unstable equilibrium; U has a local maximum.
Why it works: Negative curvature at a stationary point corresponds to a local maximum.
QUICK CHECK 3
Small-Displacement Model
Interpret the expansion.
If U(q) ≈ U0 + 1/2k(q − q0)2 with k > 0, what is the approximate generalized force near q0?
Reveal Answers
Q = −dU/dq ≈ −k(q − q0), a restoring force toward q0.
Why it works: Differentiating the quadratic energy gives a force opposite the displacement.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Equilibrium is vector balance
Both ΣF and Στ must vanish for a rigid body in static equilibrium.
KEY TAKEAWAY 2
Stability can be read from energy
A local minimum of potential energy corresponds to stable equilibrium in conservative systems.
KEY TAKEAWAY 3
Curvature controls local response
Positive energy curvature produces a restoring tendency and connects stable equilibrium to oscillatory motion.
Ready for your next step?
You've reviewed the essential ideas one last time. Now choose the resource that best matches how confident you feel. Need a quick reminder?
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