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Incline Force Mastery

A targeted mechanics lesson exploring force resolution on inclined planes and accelerating wedge systems under both constant velocity and accelerated states.

CMChuguna Murugesan

Incline Force Mastery

A targeted mechanics lesson exploring force resolution on inclined planes and accelerating wedge systems under both constant velocity and accelerated states.

CMChuguna Murugesan

Inclined Plane Forces Walkthrough Worksheet

Physics Mechanics Dynamics

Problem 1: Box on Incline with Horizontal Force \(F\)

Free-Body Diagram, Force Resolution & Net Force Equations

Name: ___________________

Date: _________

Period: _____

DIAGRAM 1: Setup & Angle of Horizontal Force \(F\) Friction is negligible • Ramp angle = \(\theta\)

θ m F θ +x' +y'

Problem Statement:

A box of mass \(m\) on an incline of angle \(\theta\) is pushed by a horizontal force \(F\). Friction between the box and incline is negligible.

Coordinates:
• \(x'\)-axis = parallel to incline (pointing up)
• \(y'\)-axis = perpendicular to incline

1. Free-Body Diagram (FBD)

Mass \(m\)

y' x'

Draw and label all forces acting on mass \(m\).

2. Force Resolution Table

Components

Force	Component along \(x'\)	Component along \(y'\)
\(F\)
\(mg\)
\(F_N\)

Resolve each force along the \(x'\) and \(y'\) axes.

Part A

Case A: Box Slides Up Incline at Constant Speed (\(a = 0\))

1. Net Force Equation along \(x'\):

\(\sum F_{x'} =\) ___________________________

Derive expression for \(F\):

\(F =\)

2. Net Force Equation along \(y'\):

\(\sum F_{y'} =\) ___________________________

Derive expression for \(F_N\):

\(F_N =\)

Part B

Case B: Box Accelerates Up Incline with Acceleration \(a\)

1. Net Force Equation along \(x'\):

\(\sum F_{x'} =\) ___________________________

Derive expression for \(F\):

\(F =\)

2. Net Force Equation along \(y'\):

\(\sum F_{y'} =\) ___________________________

Explain how acceleration affects \(F\) and \(F_N\):

Dynamics Step-by-Step Walkthrough Worksheet Page 1 of 2

Physics Mechanics Dynamics

Problem 2: Accelerating Wedge & Block System

Free-Body Diagram, Force Components & System Derivations

Page 2

DIAGRAM 2: Incline \(M\) and Block \(m\) on Frictionless Surface All surfaces frictionless

M θ m F = ?

Problem Statement:

The friction between box \(m\) and incline \(M\) and between incline \(M\) and the table are negligible.

Goal: Derive an expression for the magnitude of the horizontal force, \(F\), such that \(m\) does not slide on \(M\).

1. Free-Body Diagram of Block \(m\)

Mass \(m\)

y x

Draw and label all forces acting on block \(m\).

2. Force Resolution Table for \(m\)

Standard Axes

Force on \(m\)	Horizontal (\(x\))	Vertical (\(y\))
\(F_N\)
\(mg\)

Resolve forces on \(m\) along horizontal and vertical axes.

Case A

Case A: Block \(m\) Does NOT Slide on Incline \(M\)

1. Vertical Direction (\(y\)):

\(\sum F_y =\) ____________ \(= 0\)

Solve for \(F_N\):

\(F_N =\)

2. Horizontal Direction (\(x\)):

\(\sum F_x =\) ____________ \(= m a\)

Solve for required \(a\):

\(a =\)

3. Total System Force \(F\):

\(F =\) ______________________

Express \(F(m, M, \theta, g)\):

\(F =\)

Case B

Case B: Analysis of Constant Velocity (\(a = 0\)) & Non-Matching Acceleration

1. If the wedge moves at constant velocity (\(a = 0\)), what is the required horizontal net force on \(m\)?
\(\sum F_x =\) ________________________

2. What does this imply about normal force \(F_N\) and vertical net force \(\sum F_y\)?
__________________________________________________

Can block \(m\) remain stationary on a frictionless wedge at constant speed? [ ] Yes [ ] No

Predict the behavior of block \(m\):

• If \(F > (M+m)g\tan\theta\), block \(m\) will: ___________________________

• If \(F < (M+m)g\tan\theta\), block \(m\) will: ___________________________

• Physical reasoning:

Dynamics Step-by-Step Walkthrough Worksheet Page 2 of 2

Inclined Plane Forces Answer Key

Teacher Resource Complete Solutions

Answer Key: Problem 1 — Horizontal Push on Incline

Step-by-Step Solutions, FBD Vector Diagrams & Teaching Notes

AP / College Mechanics

Frictionless Dynamics

DIAGRAM 1: Setup & Angle Geometry of Applied Force \(F\) Answer Key Reference

θ m F θ +x' +y'

Core Geometric Point:

Because force \(F\) is horizontal, the angle between \(\vec{F}\) and the ramp surface is \(\theta\). Hence \(F_\parallel = +F\cos\theta\) and \(F_\perp = -F\sin\theta\).

Key: \(F_{\parallel} = +F\cos\theta\) • \(F_{\perp} = -F\sin\theta\)

1 Completed Free-Body Diagram

Solution

+x' +y' FN mg θ F θ

• \(F_N\) along \(+y'\) • \(F\) at angle \(\theta\) to \(+x'\) • \(mg\) at angle \(\theta\) to \(-y'\)

2 Completed Resolution Table

Solution

Force Vector	Parallel (\(x'\): up ramp)	Perp (\(y'\): ⊥ ramp)
Applied \(F\)	\(+F \cos\theta\)	\(-F \sin\theta\)
Gravity \(mg\)	\(-mg \sin\theta\)	\(-mg \cos\theta\)
Normal \(F_N\)	\(0\)	\(+F_N\)

Note: \(F\) points into the ramp (\(-y'\)); gravity pulls down-ramp (\(-x'\)).

Part A

Constant Speed Motion (\(a_{x'} = 0, \; a_{y'} = 0\)) — Solution

\(\sum F_{x'} = 0, \; \sum F_{y'} = 0\)

1. Net Force Parallel to Incline:

\(\sum F_{x'} = F \cos\theta - mg \sin\theta = 0\)

\(F \cos\theta = mg \sin\theta\)

\(F = mg \left(\frac{\sin\theta}{\cos\theta}\right)\)

Result:

\(F = mg \tan\theta\)

2. Net Force Perpendicular to Incline:

\(\sum F_{y'} = F_N - mg \cos\theta - F \sin\theta = 0\)

\(F_N = mg \cos\theta + (mg \tan\theta)\sin\theta = mg\left(\cos\theta + \frac{\sin^2\theta}{\cos\theta}\right)\)
\(= mg\left(\frac{\cos^2\theta + \sin^2\theta}{\cos\theta}\right) = \frac{mg}{\cos\theta}\)

Result:

\(F_N = \frac{mg}{\cos\theta} = mg \sec\theta\)

Part B

Accelerating Up Incline (\(a_{x'} = a > 0\)) — Solution

\(\sum F_{x'} = m a, \; \sum F_{y'} = 0\)

1. Required Applied Force \(F\):

\(\sum F_{x'} = F \cos\theta - mg \sin\theta = m a\)

\(F \cos\theta = m(g \sin\theta + a)\)

Incline Dynamics FRQ Worksheet

Physics Mechanics Dynamics • FRQ Practice

FRQ 1: Box on Rough Incline with Horizontal Force \(F\)

Kinetic Friction • Coupled Normal Force • Step-by-Step Number Substitution

Name: ___________________

Date: _________

Period: _____

θ = 30.0° m = 5.0 kg f_k F 30°

Given Numerical Parameters: Rough Ramp (\(\mu_k = 0.200\))

• Mass \(m = \mathbf{5.00\text{ kg}}\)

• Incline \(\theta = \mathbf{30.0^\circ}\)

• \(g = \mathbf{9.80\text{ m/s}^2}\)

• Friction \(\mu_k = \mathbf{0.200}\)

Key Insight: Horizontal force \(F\) presses into the ramp (\(F\sin\theta\)), increasing both \(F_N\) and friction \(f_k = \mu_k F_N\).

STEP A Calculate gravity components parallel (\(F_{g,\parallel}\)) and perpendicular (\(F_{g,\perp}\)) to the incline.

[2 Points]

Symbolic Formula & Number Substitution: Plug in \(m, g, \theta\)

Numerical Results:

\(F_{g,\parallel} =\)

N

\(F_{g,\perp} =\)

N

Include correct units

STEP B The box moves up the ramp at constant speed (\(a = 0\)). Calculate the required push force \(F\).

[4 Points]

\(\sum F_{x'} = 0\) with Friction Substitution: \(F\cos\theta - mg\sin\theta - \mu_k(mg\cos\theta + F\sin\theta) = 0\)

Required Applied Force:

\(F =\)

N

Includes overcoming \(\mu_k = 0.200\)

STEP C At constant speed, calculate the normal force \(F_N\) and the kinetic friction force \(f_k\).

[3 Points]

\(\sum F_{y'} = 0\) & \(f_k = \mu_k F_N\) Number Substitution: Plug in \(F_N = mg\cos\theta + F\sin\theta\)

Normal & Friction Forces:

\(F_N =\)

N

\(f_k =\)

N

Notice \(f_k = 0.200 \times F_N\)

STEP D The box now accelerates up the ramp at \(a = \mathbf{1.50\text{ m/s}^2}\). Calculate the new push force \(F\).

[3 Points]

\(\sum F_{x'} = ma\) Equation & Number Substitution: \(F(\cos\theta - \mu_k\sin\theta) = mg(\sin\theta + \mu_k\cos\theta) + ma\)

Accelerating Push Force:

\(F =\)

N

To 3 significant figures

Mechanics FRQ Practice • Plug-in-the-Number Calculation Sheet Page 1 of 2

Physics Mechanics Dynamics • FRQ Practice

FRQ 2: Accelerating Wedge & Block System

Coupled Dynamics • Contact Normal Force, System Acceleration & System Force \(F\)

Incline Dynamics FRQ Answer Key

Teacher Resource • Answer Key Total: 12 Points

Answer Key: FRQ 1 — Box on Rough Incline with Horizontal \(F\)

Full Number Substitutions, Kinetic Friction Analysis & Scoring Breakdown

AP Physics 1 / C Mechanics

Rough Incline • \(\mu_k = 0.200\)

θ = 30.0° m = 5.0 kg f_k F 30°

Parameters with Friction \(\mu_k = 0.200\)

\(m = 5.00\text{ kg}, \; g = 9.80\text{ m/s}^2 \implies mg = 49.0\text{ N}\)
\(\sin 30^\circ = 0.500, \; \cos 30^\circ = 0.8660\)

Key Coupling: \(F_N = mg\cos\theta + F\sin\theta\). Hence friction \(f_k = \mu_k F_N\) grows as push force \(F\) grows!

STEP A Parallel and Perpendicular Components of Gravity

Scoring: 2 pts (1 pt each)

• \(F_{g,\parallel} = mg\sin\theta = (5.00\text{ kg})(9.80\text{ m/s}^2)\sin(30.0^\circ) = (49.0\text{ N})(0.500) = \mathbf{24.5\text{ N}}\)

• \(F_{g,\perp} = mg\cos\theta = (5.00\text{ kg})(9.80\text{ m/s}^2)\cos(30.0^\circ) = (49.0\text{ N})(0.8660) = \mathbf{42.4\text{ N}}\)

Components depend only on box weight and ramp geometry.

Official Answers:

\(F_{g,\parallel} = \mathbf{24.5\text{ N}}\)

\(F_{g,\perp} = \mathbf{42.4\text{ N}}\)

Accept 42.4 N – 42.5 N

STEP B Horizontal Force \(F\) to Maintain Constant Speed with Friction

Scoring: 4 pts

\(\sum F_{x'} = F\cos\theta - mg\sin\theta - \mu_k(mg\cos\theta + F\sin\theta) = 0\) [1 pt equation]

\(F(\cos 30^\circ - 0.200\sin 30^\circ) = 49.0(\sin 30^\circ + 0.200\cos 30^\circ)\) [1 pt substitution]

\(F(0.8660 - 0.1000) = 49.0(0.500 + 0.1732) \implies F(0.7660) = 32.99\text{ N}\) [1 pt simplification]

\(F = \frac{32.99\text{ N}}{0.7660} = \mathbf{43.07\text{ N}} \approx \mathbf{43.1\text{ N}}\) [1 pt answer]

Required Force:

\(F = \mathbf{43.1\text{ N}}\)

Higher than frictionless (28.3 N)

STEP C Normal Force \(F_N\) and Kinetic Friction \(f_k\) at Constant Speed

Scoring: 3 pts

\(F_N = mg\cos\theta + F\sin\theta = 42.44\text{ N} + (43.07\text{ N})\sin(30.0^\circ)\) [1 pt]

\(F_N = 42.44 + 21.54 = \mathbf{63.98\text{ N}} \approx \mathbf{64.0\text{ N}}\) [1 pt]

\(f_k = \mu_k F_N = 0.200 \times 63.98\text{ N} = \mathbf{12.80\text{ N}} \approx \mathbf{12.8\text{ N}}\) [1 pt]

Check: \(F\cos 30^\circ = 37.3\text{ N} = F_{g,\parallel} + f_k = 24.5 + 12.8 = 37.3\text{ N}\). Balanced!

Normal & Friction:

\(F_N = \mathbf{64.0\text{ N}}\)

\(f_k = \mathbf{12.8\text{ N}}\)

Accept 63.9 N – 64.1 N

Ramp Friction Challenge Assignment

Physics Mechanics Homework Assignment

Ramp Friction Challenge: The Loading Dock Crate

Comprehensive Incline Problem • Static vs Kinetic Friction, Impending Slip & Acceleration

Name: ___________________

Date: _________

Period: _____

θ = 37.0° m = 10.0 kg T

Loading Dock Problem Statement: Total: 15 Points

A heavy crate of mass \(m = \mathbf{10.0\text{ kg}}\) rests on a rough ramp inclined at \(\theta = \mathbf{37.0^\circ}\). The surface has static coefficient \(\mu_s = \mathbf{0.500}\) and kinetic coefficient \(\mu_k = \mathbf{0.300}\). A cable pulls parallel to the ramp with tension \(T\).

\(g = 9.80\text{ m/s}^2\)

\(\sin 37.0^\circ = 0.600\)

\(\cos 37.0^\circ = 0.800\)

STEP A Unassisted Stability: Calculate \(F_{g,\parallel}\) and \(f_{s,\max}\). Will the crate slip if \(T = 0\)?

[3 Points]

Formula & Numerical Substitution: \(F_N = mg\cos\theta\) • \(f_{s,\max} = \mu_s F_N\)

Compare Values:

\(F_{g,\parallel} = \) ______ N
\(f_{s,\max} = \) ______ N

Will it slide? [ ] Yes [ ] No

STEP B Preventing Slip: Calculate minimum tension \(T_{\min}\) needed to keep the crate from sliding down.

[4 Points]

\(\sum F_\parallel = 0\) with Static Friction Directed UP the Incline: \(T_{\min} + f_{s,\max} = mg\sin\theta\)

Minimum Holding Tension:

\(T_{\min} =\)

N

Prevents downward slipping

STEP C Starting Motion Upward: Calculate minimum pull force \(T_{\text{pull}}\) to initiate upward motion.

[4 Points]

\(\sum F_\parallel = 0\) with Static Friction Directed DOWN the Incline: \(T_{\text{pull}} = mg\sin\theta + f_{s,\max}\)

Breakaway Pull Force:

\(T_{\text{pull}} =\)

N

Overcomes both gravity & friction

STEP D Active Towing: When pulled with \(T = \mathbf{125.0\text{ N}}\), find kinetic friction \(f_k\) and acceleration \(a\).

[4 Points]

\(\sum F_\parallel = ma\) Equation & Number Substitution: Use \(\mu_k = 0.300\) once crate is moving

Kinetic Friction & Accel:

\(f_k =\)

N

\(a =\)

m/s²

To 3 significant figures

Physics Dynamics Assignment • Single Incline Problem with Friction Page 1 of 1

Ramp Friction Challenge Key

Teacher Resource • Answer Key Total: 15 Points

Answer Key: Ramp Friction Challenge Assignment

Loading Dock Crate • Full Number Substitutions & Physical Analysis

AP Physics 1 / C

Static vs. Kinetic Friction

θ = 37.0° m = 10.0 kg T

Parameter Decomposition:

\(mg = (10.0\text{ kg})(9.80\text{ m/s}^2) = \mathbf{98.0\text{ N}}\)
\(F_{g,\parallel} = 98.0\sin(37.0^\circ) = \mathbf{58.8\text{ N}}\), \(F_N = 98.0\cos(37.0^\circ) = \mathbf{78.4\text{ N}}\)

Friction Limits: \(f_{s,\max} = (0.500)(78.4) = \mathbf{39.2\text{ N}}\), \(f_k = (0.300)(78.4) = \mathbf{23.52\text{ N}}\).

STEP A Unassisted Stability: Check for Impending Slip

Scoring: 3 pts

• \(F_{g,\parallel} = (10.0\text{ kg})(9.80\text{ m/s}^2)\sin(37.0^\circ) = \mathbf{58.8\text{ N}}\) [1 pt]

• \(f_{s,\max} = \mu_s mg\cos\theta = 0.500(98.0\text{ N})(0.800) = \mathbf{39.2\text{ N}}\) [1 pt]

• Comparison: \(F_{g,\parallel} (58.8\text{ N}) > f_{s,\max} (39.2\text{ N})\). Static friction cannot hold the crate alone; crate WILL SLIDE DOWN without tension. [1 pt]

Check Results:

\(F_{g,\parallel} = \mathbf{58.8\text{ N}}\)

\(f_{s,\max} = \mathbf{39.2\text{ N}}\)

✓ Will Slide Down

STEP B Minimum Tension \(T_{\min}\) to Prevent Downward Slip

Scoring: 4 pts

• Crate tends to slip downward, so static friction points UP the incline (\(+x'\)). [1 pt]

• Equilibrium: \(\sum F_\parallel = T_{\min} + f_{s,\max} - mg\sin\theta = 0\) [1 pt]

• \(T_{\min} = mg\sin\theta - f_{s,\max} = 58.8\text{ N} - 39.2\text{ N} = \mathbf{19.6\text{ N}}\) [2 pts subst & ans]

Any tension \(T \ge 19.6\text{ N}\) keeps the crate stationary.

Holding Tension:

\(T_{\min} = \mathbf{19.6\text{ N}}\)

Friction assists cable

STEP C Minimum Pull \(T_{\text{pull}}\) to Initiate Upward Motion

Scoring: 4 pts

• To pull up the ramp, static friction opposes motion and points DOWN the incline (\(-x'\)). [1 pt]

• Equilibrium at impending upward slip: \(\sum F_\parallel = T_{\text{pull}} - mg\sin\theta - f_{s,\max} = 0\) [1 pt]

• \(T_{\text{pull}} = mg\sin\theta + f_{s,\max} = 58.8\text{ N} + 39.2\text{ N} = \mathbf{98.0\text{ N}}\) [2 pts subst & ans]

Notice static equilibrium holds for any tension within \(19.6\text{ N} \le T \le 98.0\text{ N}\).

Breakaway Force:

\(T_{\text{pull}} = \mathbf{98.0\text{ N}}\)

Divide both sides by \(\cos\theta\):

Result:

\(F = \frac{m(g \sin\theta + a)}{\cos\theta}\)

2. Normal Force & Physical Behavior:

\(F_N = mg \cos\theta + F \sin\theta\)

Explanation: To accelerate faster up the ramp, \(F\) must increase. Because \(F\) has a downward component perpendicular to the ramp (\(F\sin\theta\)), a larger \(F\) presses the box harder into the incline, making \(F_N\) increase.

Teacher Answer Key • Mechanics Dynamics Walkthrough Page 1 of 2

Teacher Resource Complete Solutions

Answer Key: Problem 2 — Accelerating Wedge System

Free-Body Diagrams for Both \(m\) & \(M\), Newton's Third Law & Coupled System Dynamics

AP / College Mechanics

Coupled Bodies • Standard Axes

DIAGRAM 2: Incline \(M\) and Block \(m\) on Frictionless Surface Setup & 3rd Law Insight

M θ m F = ?

Key Coupled Interaction:

Force \(F\) acts only on \(M\). The block \(m\) accelerates solely due to the normal force \(\vec{F}_N\) from \(M\). By Newton's Third Law, \(m\) exerts an equal and opposite force \(-\vec{F}_N\) on \(M\).

Action-Reaction Pair: \(\vec{F}_{N\text{ on }m} = -\vec{F}_{N\text{ on }M}\)

1 Free-Body Diagrams: Block \(m\) & Wedge \(M\)

Completed Vector Solutions

FBD of Block \(m\) (Mass \(m\)) 2 forces

+x +y mg FN θ FNsinθ FNcosθ

• \(F_{N,x} = +F_N \sin\theta = ma\) (provides horizontal accel)

• \(F_{N,y} = +F_N \cos\theta = mg\) (supports weight, \(a_y=0\))

FBD of Wedge \(M\) (Bigger Mass) 4 forces

+x +y F Ntable Mg F'N(m on M) θ

• \(\sum F_{x,M} = F - F'_N\sin\theta = M a\) (opposed by \(m\))

• \(N_{\text{table}} = Mg + F'_N\cos\theta = (M+m)g\) (table holds all)

Case A

Equations of Motion: No Sliding of \(m\) on \(M\) (\(a_y = 0\))

Algebraic Solution

1. Vertical on Block \(m\):

\(\sum F_y = F_N \cos\theta - mg = 0\)

Vertical equilibrium holds \(m\) at constant elevation on the ramp:

\(F_N = \frac{mg}{\cos\theta}\)

2. Horizontal on Block \(m\):

\(\sum F_x = F_N \sin\theta = m a\)

Substitute \(F_N = \frac{mg}{\cos\theta}\):
\(\left(\frac{mg}{\cos\theta}\right)\sin\theta = ma\)

\(a = g \tan\theta\)

3. Total Force \(F\) on System:

\(\sum F_{x,\text{sys}} = F = (M + m) a\)

Substitute \(a = g\tan\theta\):
(Or \(F - F_N\sin\theta = Ma\))

\(F = (M+m)g \tan\theta\)

Case B

Constant Velocity (\(a = 0\)) & Force Deviation Analysis

Conceptual Solutions

1. At constant velocity (\(a = 0\)), the required horizontal net force on \(m\) is \(\sum F_x = \mathbf{0}\).

2. Since \(\sum F_x = F_N \sin\theta = 0\) and \(\theta > 0\), this requires normal force \(F_N = \mathbf{0}\).

3. With \(F_N = 0\), the vertical force is \(\sum F_y = \mathbf{-mg}\), leaving \(m\) in unsupported vertical freefall!

Can block \(m\) stay stationary at constant velocity? [ ] Yes [ ✓ ] No

Physical Predictions & Reasoning:

• If \(F > (M+m)g\tan\theta\): \(a > g\tan\theta\), so horizontal acceleration exceeds wedge slope, forcing \(F_N \cos\theta > mg\). Block \(m\) slides UP the wedge!

• If \(F < (M+m)g\tan\theta\): \(a < g\tan\theta\), so \(F_N\cos\theta < mg\). Gravity overcomes normal component, and block \(m\) slides DOWN the wedge!

Table normal force is always \(N_{\text{table}} = Mg + F_N\cos\theta = (M+m)g\) when \(m\) is in vertical equilibrium.

Teacher Answer Key • Mechanics Dynamics Walkthrough Page 2 of 2

Page 2

M = 6.0 kg 30.0° m = 2.0 kg F

Given Numerical Parameters: No Friction Anywhere

• Wedge \(M = \mathbf{6.00\text{ kg}}\)

• Block \(m = \mathbf{2.00\text{ kg}}\)

• Incline \(\theta = \mathbf{30.0^\circ}\)

• \(g = \mathbf{9.80\text{ m/s}^2}\)

Key Condition: Force \(F\) pushes \(M\) rightward such that block \(m\) does not slide on the wedge (\(a_{y,\text{block}} = 0\)).

STEP A Vertical equilibrium on \(m\): Calculate normal force \(F_N\) exerted by wedge \(M\) on block \(m\).

[2 Points]

\(\sum F_y = 0\) Equation & Number Substitution: Notice \(F_{N,y} = F_N\cos\theta = mg\)

Normal Contact Force:

\(F_N =\)

N

To 3 significant figures

STEP B Horizontal motion of \(m\): Calculate the required rightward acceleration \(a\) of the block.

[3 Points]

\(\sum F_x = ma\) Equation & Number Substitution: \(F_N\sin\theta = ma \implies a = g\tan\theta\)

Required Acceleration:

\(a =\)

m/s²

Independent of mass \(m\)

STEP C Total system: Calculate the horizontal force \(F\) pushing the wedge to achieve this acceleration.

[3 Points]

System \(F = (M+m)a\) Equation & Number Substitution: Plug in \((M+m)\) and \(a\) from Step B

Required System Push \(F\):

\(F =\)

N

To 3 significant figures

STEP D Calculate normal force \(N_{\text{table}}\) from the floor on wedge \(M\); predict motion if \(F\) drops to \(0\).

[4 Points]

Calculation of \(N_{\text{table}}\) & Physical Reasoning when \(F = 0\): Show \(N_{\text{table}} = Mg + F_N\cos\theta\)

Floor Normal & Behavior:

\(N_{\text{table}} =\)

N

If \(F = 0\), block \(m\) will:

[ ] Slide Up [ ] Slide Down

Mechanics FRQ Practice • Plug-in-the-Number Calculation Sheet Page 2 of 2

STEP D Increased Force \(F\) to Produce Acceleration \(a = 1.50\text{ m/s}^2\)

Scoring: 3 pts

\(\sum F_{x'} = F(\cos\theta - \mu_k\sin\theta) - mg(\sin\theta + \mu_k\cos\theta) = ma\) [1 pt]

\(F(0.7660) - 32.99\text{ N} = (5.00\text{ kg})(1.50\text{ m/s}^2) = 7.50\text{ N}\) [1 pt]

\(F = \frac{32.99 + 7.50}{0.7660} = \frac{40.49\text{ N}}{0.7660} = \mathbf{52.86\text{ N}} \approx \mathbf{52.9\text{ N}}\) [1 pt]

New \(F_N = 42.44 + (52.86)(0.500) = \mathbf{68.9\text{ N}}\), new \(f_k = \mathbf{13.8\text{ N}}\).

Accelerating Force:

\(F = \mathbf{52.9\text{ N}}\)

+9.8 N above constant-speed \(F\)

Teacher Answer Key • Incline Dynamics FRQ Practice Page 1 of 2

Teacher Resource • Answer Key Total: 12 Points

Answer Key: FRQ 2 — Accelerating Wedge & Block System

Contact Force Resolution, System Acceleration & Floor Normal Solutions

AP Physics 1 / C Mechanics

Coupled Bodies • Standard Axes

M = 6.0 kg 30.0° m = 2.0 kg F

Given Parameters & Values Total System = 8.00 kg

\(M = 6.00\text{ kg}, \; m = 2.00\text{ kg}, \; \theta = 30.0^\circ, \; g = 9.80\text{ m/s}^2\)
Weight of block: \(mg = (2.00)(9.80) = \mathbf{19.6\text{ N}}\)

Key Concept: Normal force \(F_N\) provides BOTH vertical lift (\(F_N\cos\theta = mg\)) and horizontal acceleration (\(F_N\sin\theta = ma\)).

STEP A Vertical Equilibrium: Calculate Normal Force \(F_N\) on Block \(m\)

Scoring: 2 pts

\(\sum F_y = F_N\cos\theta - mg = 0 \implies F_N\cos(30.0^\circ) = (2.00\text{ kg})(9.80\text{ m/s}^2) = 19.6\text{ N}\) [1 pt]

\(F_N = \frac{19.6\text{ N}}{\cos(30.0^\circ)} = \frac{19.6\text{ N}}{0.8660} = \mathbf{22.63\text{ N}} \approx \mathbf{22.6\text{ N}}\) [1 pt]

Angle between \(\vec{F}_N\) and vertical is \(\theta = 30^\circ\). Therefore the vertical lift is \(F_N\cos\theta\).

Normal Force on \(m\):

\(F_N = \mathbf{22.6\text{ N}}\)

Accept 22.6 N – 22.7 N

STEP B Horizontal Motion: Calculate System Acceleration \(a\)

Scoring: 3 pts

\(\sum F_x = F_N\sin\theta = ma \implies (22.63\text{ N})\sin(30.0^\circ) = (2.00\text{ kg})a\) [1 pt]

\(11.31\text{ N} = (2.00\text{ kg})a \implies a = \frac{11.31}{2.00} = \mathbf{5.66\text{ m/s}^2}\) [1 pt]

Or analytically: \(a = g\tan\theta = (9.80)\tan(30.0^\circ) = \mathbf{5.66\text{ m/s}^2}\). Notice this is completely independent of mass \(m\)! [1 pt]

Acceleration:

\(a = \mathbf{5.66\text{ m/s}^2}\)

Units (m/s²) required

STEP C Total System: Calculate Horizontal Applied Force \(F\)

Scoring: 3 pts

System equation: \(F = (M + m)a = (6.00\text{ kg} + 2.00\text{ kg})a\) [1 pt]

\(F = (8.00\text{ kg})(5.658\text{ m/s}^2) = \mathbf{45.26\text{ N}} \approx \mathbf{45.3\text{ N}}\) [1 pt]

Check via wedge FBD: \(F - F_N\sin\theta = Ma \implies F = (6.00)(5.658) + 11.31 = 33.95 + 11.31 = 45.3\text{ N}\). [1 pt]

Total Push Force:

\(F = \mathbf{45.3\text{ N}}\)

Accept 45.2 N – 45.3 N

STEP D Table Normal Force \(N_{\text{table}}\) & Behavior when \(F = 0\)

Scoring: 4 pts

Vertical equilibrium of entire system: \(\sum F_{y,\text{sys}} = N_{\text{table}} - (M+m)g = 0\) [1 pt]

\(N_{\text{table}} = (6.00 + 2.00)(9.80) = (8.00\text{ kg})(9.80\text{ m/s}^2) = \mathbf{78.4\text{ N}}\) [1 pt]

(Also via wedge FBD: \(N_{\text{table}} = Mg + F_N\cos\theta = 58.8 + 19.6 = 78.4\text{ N}\)). [1 pt]

When \(F = 0\): Wedge cannot accelerate rightward to support the block; \(F_N\cos\theta < mg\), so net vertical force is downward: block Slides Down the wedge. [1 pt]

Floor Normal & Motion:

\(N_{\text{table}} = \mathbf{78.4\text{ N}}\)

✓ Slide Down

Gravity overcomes normal lift

Teacher Answer Key • Incline Dynamics FRQ Practice Page 2 of 2

Cable fights gravity + friction

STEP D Active Towing: Kinetic Friction \(f_k\) & Acceleration \(a\) (\(T = 125.0\text{ N}\))

Scoring: 4 pts

• Kinetic friction while moving: \(f_k = \mu_k F_N = 0.300(78.4\text{ N}) = \mathbf{23.52\text{ N}} \approx \mathbf{23.5\text{ N}}\) [1 pt]

• Newton's 2nd Law: \(\sum F_\parallel = T - mg\sin\theta - f_k = ma\) [1 pt]

• \(125.0\text{ N} - 58.8\text{ N} - 23.52\text{ N} = 42.68\text{ N} = (10.0\text{ kg})a\) [1 pt]

• \(a = \frac{42.68\text{ N}}{10.0\text{ kg}} = \mathbf{4.268\text{ m/s}^2} \approx \mathbf{4.27\text{ m/s}^2}\) [1 pt]

Active Motion:

\(f_k = \mathbf{23.5\text{ N}}\)

\(a = \mathbf{4.27\text{ m/s}^2}\)

Accept 4.26 – 4.27 m/s²

Teacher Answer Key • Ramp Friction Challenge Assignment Page 1 of 1

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