Show Me The Physics - Static Equilibrium

Intro to Finding Resultants

Pre-Class Questions 1 - 3

▶ Instruction Video [1] ▶ Instruction Video [2]

Review – Finding Resultants ($F_{net}$) $0^\circ$, $180^\circ$, $90^\circ$

Defined - Resultant - Single force equivalent to all vectors acting on an object.

1. Angle between vectors below = ______________

$F_{net} = $

$F_{net} = $

2. Angle between vectors below = ______________

$F_{net} = $

Teacher Answer Key

Physics Equation: $F_{net} = \sum F$ (Vector addition based on direction)

Claim-Evidence-Reasoning: Claim: The resultant magnitude is at its maximum ($R = A + B$) at $0^\circ$ and minimum ($R = |A - B|$) at $180^\circ$.
Evidence: For the $0^\circ$ case, $F_{net} = 5.0\text{ N} + 7.0\text{ N} = 12.0\text{ N}$. For the $180^\circ$ case, $F_{net} = 5.0\text{ N} - 4.0\text{ N} = 1.0\text{ N}$.
Reasoning: Vectors that point in the same direction add their total effect to the system, while opposing vectors ($180^\circ$ apart) subtract their magnitudes because they work against each other.

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Naming Direction with Angles

3. The vector on the upper right is 50. degrees N of E. Name the rest of the directions.

________________________

4. Angle between vectors = 90. degrees. Use TRIG to find the direction.

What trig. function helps you find the $F_{net}$ direction? ________

$F_{net}$ = _____ ___ degrees _____ of _______

5. 10. nt east and 10. nt south acts on an object. Find $F_{net}$.

Calculation: _____________________________________________________

$F_{net}$ = _____ ___ degrees _____ of _______

Teacher Answer Key

Physics Equations: $c = \sqrt{a^2 + b^2}$ (Pythagorean Theorem) and $\theta = \tan^{-1}(\frac{\text{Opposite}}{\text{Adjacent}})$

Problem 5 Math: $F_{net} = \sqrt{(10.0\text{ N})^2 + (10.0\text{ N})^2} = 14.1\text{ N}$ at $45^\circ$ South of East.

Claim-Evidence-Reasoning: Claim: Vectors acting at $90^\circ$ produce a resultant that is the hypotenuse of a right triangle.
Evidence: For two $10.0\text{ N}$ perpendicular forces, the math yields a resultant of $14.1\text{ N}$.
Reasoning: When two forces are perpendicular, they act independently along their axes. Their combined effect is found by the Pythagorean theorem, representing the shortest distance between the start of the first vector and end of the last.

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Finding $F_{net}$ Using Components

6. In AP Physics it is more useful to find $F_{net}$ using components.

Vector	X Component (N)	Y Component (N)
Vector A
Vector B
$F_{net}$

1. Find Components of all vectors (Use neg. and pos.)

2. Add Components

3. _________________________ to get the magnitude

4. ________________________ to get the direction

$F_{net} = $ _________ N _______ degrees _______ of ____

7. Find $F_{net}$ By Adding Components and using Tangent.

8. Which pair of vectors has the greatest resultant?

What’s the TAKEAWAY Here? ______________________________

9. Relative to the ground, an airplane gains speed with wind from behind and loses speed head-on. When wind is at a right angle, ground speed:

A. increases. B. decreases. C. same. D. need info.

Teacher Answer Key

Physics Equations: $F_x = F\cos\theta$, $F_y = F\sin\theta$, $c = \sqrt{a^2+b^2}$

Claim-Evidence-Reasoning: Claim: Resultant forces are maximized when vectors act in the same direction ($0^\circ$) and increase ground speed when air velocity and wind act as perpendicular components.
Evidence: Calculations for Q8 show the $0^\circ$ pair yields the highest magnitude. In Q9, the ground speed is the hypotenuse ($V_g = \sqrt{V_a^2 + V_w^2}$).
Reasoning: Perpendicular vectors form a right triangle where the hypotenuse is always longer than either leg. Therefore, a $90^\circ$ crosswind increases the magnitude of the ground speed vector compared to the airspeed alone.

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Free Body Diagrams – Static Equilibrium

Newton’s 2nd Law is NOT …. $F = ma$

Static Equilibrium means: 1. __________ 2. __________ 3. __________

A. Most of the time it is useful to write: $\sum F_x = $ ________ and $\sum F_y = $ ________

Example Problems

1. Show all the forces acting on the body below each picture. 2. Show Equation and Solve.

FBD

1. Show forces 2. Solve:

Solve these without a Free Body Diagram:

FBD with Mass $m$

7. An orb of mass 5.0 kg is hanging from the ceiling. Find the tension of the string ($T$)

8. If mass $M$ is at rest on a rough table, find a) tension $T$ b) $M$ in terms of $m$

a) $T$ = ____________ b) $M$ = ____________________

Teacher Answer Key

Physics Equations: $\sum F = 0\text{ N}$ (Equilibrium), $F_g = mg$, $T - F_g = 0\text{ N}$

Problem 7 Math: $T = (5.0\text{ kg})(10.0\text{ m/s}^2) = 50.0\text{ N}$.

Claim-Evidence-Reasoning: Claim: In static equilibrium, the net force acting on the object is $0\text{ N}$.
Evidence: For the $5.0\text{ kg}$ hanging orb, the tension is measured at $50.0\text{ N}$.
Reasoning: Gravity pulls the mass down with a force of $mg$. Since the object is at rest, Newton's 1st Law states that an equal and opposite force (Tension) must balance it exactly so that $F_{net} = 0\text{ N}$.

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Free Body Diagrams Statics – Acute Angles

1. a) Draw FBD b) Show ALL Equations c) Solve unknowns

3. a) Draw FBD b) Show ALL Equations c) Solve unknowns

Teacher Answer Key

Physics Equations: $\sum F_y = T_{1y} + T_{2y} - F_g = 0\text{ N}$ and $T_y = T\sin\theta$

Math (Symmetric 10 kg mass at 30°): $2T\sin(30^\circ) = 100.0\text{ N} \implies 2T(0.5) = 100.0\text{ N} \implies T = 100.0\text{ N}$.

Claim-Evidence-Reasoning: Claim: Tension in supporting strings increases significantly as the angle with the horizontal decreases.
Evidence: To support a $100.0\text{ N}$ weight with two strings at $30^\circ$, each string must exert $100.0\text{ N}$ of tension.
Reasoning: Only the vertical components of tension ($T\sin\theta$) counteract gravity. As the angle becomes shallower, each string must pull with much more total force just to maintain the required vertical component to balance the weight.

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Inclined Planes – Free Body Diagrams

Interactive Vector Simulation

Ramp Angle: 30°

1. The force (N) the box exerts ON the ramp is: $F_{\perp}$ OR $F_{||}$? (Circle)

2. What ramp angle ($^\circ$) causes force on ramp to be: a) Greatest? ____ b) Smallest? ____

3. At what angle ($^\circ$) is $F_{on-ramp} = $ Parallel Component of weight? ___________

4. What ramp angle ($^\circ$) would $F_N = F_g$? _______

5. The larger the ramp angle, the SMALLER or LARGER the Normal Force (circle)

Frictionless Ramp / Trig and Weight Components

Ex 1) 5.0 kg box slides on 20. degree frictionless ramp.

Ex 2) Ball mass = 5.0 kg and angle = 45 degrees.

Ex 4) Mass 5.0 kg, angle = 70 degrees.

Ex 5) Mass 5.0 kg at steep angle.

Ex 6) Mass 5.0 kg at shallow angle.

Static Equilibrium – Incline Review

Parallel Component = __________________

Perpendicular Component = _______________

Part I - Pencil Tasks

Label the ramp angle $\theta$.
Draw perpendicular and parallel components of $F_g$.
Label components in terms of $F_g$ and $\theta$.
Draw the normal force on the box.

Part II - Analysis

1. What happens to the magnitude of an object's parallel and perpendicular components on a ramp when the angle is increased?

____________________________________________________________

2. At what ramp angle is the parallel component of a box on a ramp at its minimum magnitude?

____________________________________________________________

Teacher Answer Key

Physics Equations: $F_{||} = mg\sin\theta$ and $F_{\perp} = mg\cos\theta$

Claim-Evidence-Reasoning: Claim: As ramp angle increases, the parallel force component increases while the perpendicular force component decreases.
Evidence: Trig functions show $\sin(0^\circ) = 0.0$ and $\cos(0^\circ) = 1.0$. At $90^\circ$, $\sin(90^\circ) = 1.0$ and $\cos(90^\circ) = 0.0$.
Reasoning: $F_{||}$ depends on the sine of the angle; as the ramp gets steeper, more of gravity's pull is directed along the surface. $F_{\perp}$ depends on cosine; as the ramp gets steeper, the object "squeezes" the ramp less until it is vertical and $F_{\perp}$ becomes $0.0\text{ N}$.

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Worksheet: Incline Plane Problems

1. The box on a frictionless ramp is held at rest by tension. Mass = 20. kg. Angle = 30°. Find $T$.

2. Pulley and ramp are frictionless, block in static equilibrium. Find mass $m$.

3. Ed is sledding down a 15 degree hill. Mass = 54 kg. What force pulls them down?

4. Brick held on frictionless incline by tension. Find $T$ in terms of $m$.

Challenge: Brick at rest on smooth ramp attached to hanging mass. Solve for $M$ in terms of $m$.

Forces on Inclined Planes: Additional Practice

1. While being unloaded, a 10. kg box is placed on a 37 degree ramp. Box does not move. Find $\mu_s$.

2. $\mu_s = 0.25$. Find minimum angle required for sliding.

Teacher Answer Key

Physics Equations: $F_{||} = F_f$ (at threshold) and $F_f = \mu F_N$

Math Derivation: $mg\sin\theta = \mu mg\cos\theta \implies \mu = \tan\theta$

Problem 1 Math: $\mu_s = \tan(37.0^\circ) \approx 0.75$.

Problem 2 Math: $\theta = \tan^{-1}(0.25) \approx 14.0^\circ$.

Claim-Evidence-Reasoning: Claim: The coefficient of static friction is directly determined by the tangent of the slope's angle at the limit of stability.
Evidence: Mathematical cancellation shows that $mg$ drops out of the equation, leaving $\mu = \tan\theta$.
Reasoning: At the slip point, the component of gravity pulling the object down the ramp ($mg\sin\theta$) is perfectly balanced by the frictional force pushing up ($\mu mg\cos\theta$). This ratio is independent of the object's mass.

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