Calculation of Required Force – Complete Notes, Revision, Important Questions & Downloads
Calculation of Required Force covers two TOC subtopics: Pull / Push at Angle on Horizontal Surface and Minimum Force. NEET tests: (1) pull formula P = μmg/(cosα + μ sinα), (2) push formula, (3) minimum force P_min = μmg/√(1+μ²) at angle = angle of friction. Six canonical scenarios where force P is applied at various angles on horizontal surfaces: pulling up at angle α above horizontal — P = μ(mg − P sinα)/cosα; pushing down at angle α below — P = μ(mg + P sinα)/cosα. The minimum force result: P_min = μmg/√(1+μ²) = mg sinθ when applied at φ = θ = angle of friction. 'For minimum value of P its angle from the horizontal should be equal to angle of friction.'
NEET Weightage — Calculation of Required Force
Friction (Chapter 5)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 1 | 4 | |
| 2022 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2020 | 0 | 0 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019–2024) | 2–4 | 8–16 |
MINIMUM FORCE ON HORIZONTAL SURFACE: For pulling at angle α: P = μmg/(cosα + μ sinα). To minimise P: maximise (cosα + μ sinα). Differentiate: d/dα(cosα + μ sinα) = −sinα + μ cosα = 0 → tanα = μ = μ_s → α = tan⁻¹(μ) = angle of friction. NCERT: 'For minimum value of P its angle from the horizontal should be equal to angle of friction.' P_min at α = θ (angle of friction): P_min = μmg/(cosθ + μ sinθ). Using tanθ = μ: sinθ = μ/√(1+μ²), cosθ = 1/√(1+μ²). P_min = μmg/(1/√(1+μ²) + μ²/√(1+μ²)) = μmg/((1+μ²)/√(1+μ²)) = μmg/√(1+μ²) = mg sinθ.
INCLINED SURFACE SCENARIOS: Pulling UP an incline (angle α_inc, force P at angle β to incline): gravity component down the incline = mg sinα_inc. Friction acts DOWN the incline (opposing upward motion). At onset: P = (mg sinα_inc + μ mg cosα_inc) when P is along the incline (β = 0). For best angle, similar calculus as horizontal case. PUSHING DOWN the incline: friction acts up the incline. At onset: P = mg sinα_inc − μ mg cosα_inc (assuming sinα_inc > μ cosα_inc, i.e., incline > angle of repose). NEET focuses on scenarios 1, 2, and minimum force.
How to Prepare Calculation of Required Force for NEET
Master the force resolution for pull vs push scenarios Always draw forces: weight mg (down), normal R (perpendicular to surface), applied force P (at angle α), friction f = μR (along surface, opposing motion). Key difference: PULL at angle α above horizontal → R = mg − P sinα (reduced). PUSH at angle α below horizontal → R = mg + P sinα (increased). This is the single most important distinction. Wrong sign → wrong answer. For inclined surfaces: also add/subtract gravity component along incline to friction condition.
Memorise the minimum force formula and the optimal angle P_min = μmg/√(1+μ²). Optimal angle = angle of friction θ = tan⁻¹(μ). Also express: P_min = mg sinθ. NEET may give μ and ask for P_min: use P_min = μmg/√(1+μ²). Or give θ and ask: use P_min = mg sinθ. Practice both directions. Derivation shortcut: at optimal angle, P = μmg/(cosθ + μ sinθ) — substitute cosθ = 1/√(1+μ²) and sinθ = μ/√(1+μ²).
Practise all six scenarios as quick resolved-force setups Scenario 1 (pull, horizontal): P = μmg/(cosα+μsinα). Scenario 2 (push, horizontal): P = μmg/(cosα–μsinα). Scenario 3 (up incline, force along incline): P = mg(sinθ+μcosθ). Scenario 4 (down incline, force along incline): P = mg(sinθ–μcosθ). Scenario 5 (push down to prevent motion): P = mg(μcosθ–sinθ). Scenario 6 (min force, horizontal): P_min = μmg/√(1+μ²) at angle θ = tan⁻¹μ.
Study Materials — Calculation of Required Force
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Rapid Revision — Calculation of Required Force
Concept → Trap → Example1) Pull / Push at Angle on Horizontal Surface
Pull / Push at Angle on Horizontal SurfaceSETUP: a block of mass m is on a horizontal rough surface (μ = coefficient of friction). A force P is applied at angle α to move the block. Two fundamentally different cases: CASE 1 — PULL (force inclined ABOVE horizontal at angle α): The vertical component P sinα acts UPWARD (opposite to mg). Normal reaction: R = mg − P sinα (REDUCED compared to P horizontal). Friction: f = μR = μ(mg − P sinα). Horizontal condition at onset of motion: P cosα = μ(mg − P sinα). Rearranging: P cosα + μP sinα = μmg → P(cosα + μ sinα) = μmg → P_pull = μmg/(cosα + μ sinα). Key insight: pulling at angle α reduces normal force → reduces friction → pulling requires LESS force compared to pushing at same angle. CASE 2 — PUSH (force inclined BELOW horizontal at angle α): The vertical component P sinα acts DOWNWARD (same direction as mg). Normal reaction: R = mg + P sinα (INCREASED). Friction: f = μR = μ(mg + P sinα). Horizontal condition at onset: P cosα = μ(mg + P sinα) → P(cosα − μ sinα) = μmg → P_push = μmg/(cosα − μ sinα). Key insight: pushing at angle increases normal force → increases friction → more force needed. Also: cosα − μ sinα could be small or zero if tanα = 1/μ → P → ∞ (impossible to slide by pushing at this angle — effectively locked). INCLINED PLANE (force along incline, incline angle φ): UP the incline: P = mg(sinφ + μ cosφ). DOWN the incline (opposing slide down): P = mg(sinφ − μ cosφ) if sin φ > μ cos φ.
- COMPARISON: Pull vs Push at same angle α: P_pull = μmg/(cosα + μ sinα) and P_push = μmg/(cosα − μ sinα). Since (cosα + μ sinα) > (cosα − μ sinα) for μ > 0, α > 0: denominator for pull is LARGER → P_pull < P_push. Pulling always requires less force than pushing at the same angle magnitude. This is a direct NEET multiple-choice concept: 'It is easier to pull a lawn roller than to push it' — pulling reduces normal force and friction; pushing increases both.
- SPECIAL CASE — pulling horizontally (α = 0): P = μmg/(cos0 + μ sin0) = μmg/1 = μmg. This is the familiar formula — force required to slide horizontally = μmg. For pulling at α > 0: P < μmg (less force needed). For pushing at α > 0: P > μmg (more force needed). The minimum of P_pull occurs at optimal angle α = θ (angle of friction, see Minimum Force revision card). For P_push: P increases monotonically with α (no minimum, only increases as α increases).
- INCLINED SURFACE SCENARIOS: On incline of angle φ: (i) Force P applied UP the incline (along incline surface): Driving up requires overcoming gravity component AND kinetic friction: P = mg sinφ + μ_k mg cosφ. (ii) Force P to hold body on incline from sliding DOWN: P (up the incline) + f (up the incline) = mg sinφ, so P = mg sinφ − f = mg sinφ − μ_s mg cosφ (at limiting condition). If φ < angle of repose: no force needed (body stays by itself). (iii) Force P pushing body DOWN the incline: gravity helps; friction opposes; P = mg sinφ − μ_k mg cosφ if already sliding. Setting up the correct sign of friction (direction of motion determines friction direction) is critical.
2) Minimum Force — P_min = μmg/√(1+μ²)
Minimum ForcePROBLEM: find the angle α at which the applied force P to slide a block on a horizontal surface is MINIMUM. STARTING FORMULA: P_pull(α) = μmg/(cosα + μ sinα). To minimise P: maximise the denominator f(α) = cosα + μ sinα. METHOD 1 (Calculus): d/dα (cosα + μ sinα) = −sinα + μ cosα = 0 → sinα/cosα = μ → tanα = μ → α = tan⁻¹(μ) = angle of friction θ. METHOD 2 (Trigonometric): cosα + μ sinα = R sin(α + φ) form. cosα + μ sinα = √(1+μ²) × cos(α − tan⁻¹(μ)). Maximum when cos(α − tan⁻¹(μ)) = 1, i.e., α = tan⁻¹(μ). NCERT: 'For minimum value of P its angle from the horizontal should be equal to angle of friction. For the force P to be minimum (cosα + μ sinα) must be maximum.' MINIMUM FORCE VALUE: at α = θ (angle of friction), sinθ = μ/√(1+μ²), cosθ = 1/√(1+μ²). P_min = μmg/(cosθ + μ sinθ) = μmg/(1/√(1+μ²) + μ²/√(1+μ²)) = μmg × √(1+μ²)/(1+μ²) = μmg/√(1+μ²). ALTERNATIVE FORM: P_min = mg sinθ (since μ/√(1+μ²) = sinθ). SUMMARY: Minimum force = μmg/√(1+μ²) = mg sinθ, applied at angle θ = tan⁻¹(μ) above horizontal.
- NUMERICAL EXAMPLES: μ = 1 (θ = 45°): P_min = mg/√2 ≈ 0.707mg. At horizontal (α=0): P = μmg = mg. Minimum is significantly less (0.707mg vs mg) — 30% less force at 45°. For μ = 0.5 (θ ≈ 26.6°): P_min = 0.5mg/√(1+0.25) = 0.5mg/√1.25 = 0.5mg/(√5/2) = mg/√5 ≈ 0.447mg. At horizontal: P = 0.5mg. Min is 0.447mg vs 0.5mg — 10.6% less. The benefit of applying at optimal angle increases as μ increases.
- WHY OPTIMAL ANGLE = ANGLE OF FRICTION: At α = θ (angle of friction), the applied force P, normal reaction R, and the limiting friction F_l form a specific geometric relationship. At this angle, P is perpendicular to the resultant contact force S = √(F_l² + R²). The applied force exactly overcomes the contact force in the most efficient direction. Geometrically: P is perpendicular to S → P is tangent to the friction circle — minimum P to push outside the friction cone. This is the physical reason: applying P perpendicular to the resultant contact force requires the least magnitude.
- NEET EXAM TIPS: Two most common question types: (1) Given μ (or θ), find P_min — use P_min = μmg/√(1+μ²). (2) At what angle should force be applied for minimum force? — Answer: angle of friction = tan⁻¹(μ). Also check: is P_min > P_horizontal? No — P_min is ALWAYS ≤ μmg (the horizontal value). At α = θ: P = P_min. Applying P at any other angle gives a larger force requirement. Pushing at α < 0 (downward) always requires MORE than μmg.
US Curriculum Gaps — Calculation of Required Force
Topics in this section are tested in NEET but may be organised differently in US physics courses.Six-Scenario Force Enumeration in AP Physics 1
AP Physics 1 requires students to set up force equations for blocks on inclined surfaces and horizontal surfaces with applied forces at angles. However, NCERT explicitly enumerates six canonical force-calculation scenarios as a structured set, and NEET tests them as a distinct topic category. AP students learn the physics via free-body diagrams for individual problems but may not see the systematic comparison of pull vs push at angle (and the key insight that pull at angle reduces normal force while push increases it). The six-scenario framework is an NCERT-specific pedagogical structure valuable for NEET.
- AP Physics 1: solves individual force-angle problems; no named 6-scenario enumeration
- NEET: all 6 scenarios (pull/push horizontal, pull/push incline up/down, minimum force) are testable
- Key NEET insight: pull at angle reduces normal force (R = mg − P sinα); push increases it (R = mg + P sinα)
Minimum Force Derivation and Angle-of-Friction Optimality in University Physics
The derivation that minimum force to slide a block occurs at the angle of friction (via calculus or trigonometric maximisation of cosα + μ sinα) appears in NCERT as a named result. In Halliday & Resnick and most AP Physics 1 materials, this optimisation may appear as an exercise but is not a named result. NEET tests both the formula (P_min = μmg/√(1+μ²)) and the concept (optimal angle = angle of friction). US students comfortable with trigonometric identity maximisation (R sin(α + φ) form) will find the derivation straightforward, but should explicitly practise the formula P_min = μmg/√(1+μ²).
- NEET: P_min = μmg/√(1+μ²) is an expected formula; optimal angle = angle of friction is a stated rule
- AP Physics 1: optimisation of applied force angle not a standard named result
- US Halliday: may appear as calculus exercise; NEET tests it as a formula-level result
NEET-Style Practice Questions — Calculation of Required Force
4 QuestionsPractice Problems — Calculation of Required Force
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Physics — Friction Revision Checklist
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Calculation of Required Force
Notes · Downloads · Revision · Important QuestionsWhy does pulling at an angle above horizontal require less force than pulling horizontally?
What is the minimum force to slide a block on a rough surface, and at what angle should it be applied?
Why is the optimal angle for minimum force equal to the angle of friction?
What happens to the required force if you push the block at an angle instead of pulling?
How do I set up the force equation for pushing up an inclined surface?
For what angle does P_push become infinite (block cannot be slid by pushing)?
Is P_min always less than the force required at 0° (horizontal)?
Does the mass of the block affect the optimal angle for minimum force?
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