Newton’s 2nd Law

Newton's 2nd Law Header

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1st and 2nd Law Notes / Worksheet

A) 3 Laws of Motion - 1st Law - Law of Inertia Video Instruction

If $\Sigma F_{net} = 0$, then $a = $

Because

$\Sigma F_{net} = 0$ means that the object is at rest or moving with

Ex) Free Body Diagram - Block being pulled to the right at a constant velocity.

Draw all the forces:

Constant Velocity FBD
Teacher Solution: CER Method

Claim $a = 0 \text{ m/s}^2$. The forces are balanced in all directions.

Evidence Constant velocity implies $\Delta v = 0$, therefore $a = 0$. Since $\Sigma F = ma$, the net force is $0 \text{ N}$.

Reasoning Per Newton’s 1st Law, a constant velocity is a state of equilibrium. The applied force to the right must be exactly equaled by the force of friction to the left.

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A net force is not required

A 5.0 kg box is moved across a floor at constant velocity with a horizontal F of 12 N. What is the force of friction on the box?

2nd Law - Relationship between F, m, a

$\Sigma F_{net} = $

Unbalanced forces cause objects to

$F_{net} = \text{Resultant}$

Vertical Lift Application

Ex) Mass M below is lifted up with an acceleration $a$. Find the tension in the string T

Vertical Lift Diagram

$\Sigma F_{net} = ma$ (up )

Teacher Solution: CER Method

Claim Tension $T = M(g + a)$.

Evidence $\Sigma F_y = T - F_g = Ma$.
$T - M(10 \text{ m/s}^2) = M(a) \implies T = M(10 + a)$.

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If the String is Accelerated downward?

Connected Masses

Two masses are connected by a string. Video Instruction p. 3 - 5

Ex 1) Two masses are connected by a string and remain at rest. Find tension force on top and bottom rope.

Two Connected Masses at Rest

$\Sigma F = $

Top String Tension Force:

Bottom String Tension Force:

Teacher Solution: CER Method

Claim $T_{bottom} = m_B (10 \text{ m/s}^2)$ and $T_{top} = (m_A + m_B)(10 \text{ m/s}^2)$.

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Ex 2) Find the tension of the top and bottom string if objects accelerated upward.

Connected Masses Accelerated

Free body diagram B (easier) Sketch all forces on B

Sketch all forces on A

Cart Motion Analysis

Cart Motion Comparison

Which cart is slowing down as it travels to the right?

Describe the motion of the rest of the cars.

Teacher Solution: CER Method

Claim Cart C is slowing down.

Reasoning The cart travels right, but the net force (acceleration) points left. When velocity and acceleration are opposite, the object slows down.

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Ramp / F = ma

Resolving Weight force on Ramp

Incline Force Resolution

Label $F_g \sin \theta$ and $F_g \cos \theta$ (Parallel & perpendicular components)

Find (f) friction if the box is stationary. CAR at rest, Find $\mu_s$

Teacher Solution: CER Method

Claim $\mu_s = \tan \theta$.

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The box rests on a ramp. What is T in terms of M, $\theta$

Box Equilibrium on Ramp

Weights on a Pulley

Video Instruction

Assuming a frictionless, massless pulley, ... determine the acceleration of the blocks after the large block is released.

Atwood Machine

Force of the tension of the rope on mass m, M?

Teacher Solution: CER Method

Claim $a = \frac{(M - m) \cdot 10}{M + m}$. Rule: One Tension.

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Box / Frictionless Pulley / Table

Video Instruction

Ex 1) Find acceleration for this system (frictionless table)

Table Pulley System

Two bodies move

$F_{net} = $

Teacher Solution: CER Method

Claim $a = \frac{m(10)}{M+m}$.

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Ex 2) Acceleration if smaller mass was hanging?

Massless, Frictionless Pulley and Ramp

Find the acceleration of the red box up the ramp (M)

Ramp Pulley Diagram

Masses move

  1. When does a = 0?
  2. b) Under what circumstances will the large box accelerate down the ramp?
Teacher Solution: CER Method

Claim $a=0$ if $m_{hang} = M_{ramp} \sin \theta$.

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Second Law - $F_{net} = ma$ / No Friction

Finding Normal Force / Acceleration

Incline Diagram Start

$F_{net} = $

$m = $

Vector Components
Teacher Solution: CER Method

Claim $a = (10 \text{ m/s}^2) \sin \theta$.

Reasoning Gravity is the only unbalanced force. Mass cancels out.

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DYNAMICS $\Sigma F \neq 0$ Word Problems

1) A 10.-kg cement block is pulled to the right across a rough floor (μ≠0) by a horizontal force of 40. N. If the cement block accelerates to the right at 1.0 m/s2, what is the μ between the floor and the cement block?

Teacher Solution: CER Method

Claim $\mu = 0.30$.

Evidence $40 \text{ N} - f = (10 \text{ kg})(1.0 \text{ m/s}^2) \implies f = 30 \text{ N}$. Normal Force $F_N = 100 \text{ N}$. $\mu = 30 / 100 = 0.30$.

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2) A horizontal force pushes a 20.-kg box to the right across a rough floor where μ = 0.50. If the box accelerates to the right at 6.0 m/s2, what is the magnitude of the horizontal push?

Teacher Solution: CER Method

Claim $F_{push} = 220 \text{ N}$.

Evidence $f = (0.5)(200) = 100 \text{ N}$. $F_{net} = ma = 120 \text{ N}$. $F_{push} = 120 + 100 = 220 \text{ N}$.

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3) A 40.0 -kg box is pulled to the right across a rough floor (μ≠0) by a rope angled at 36.87° above the horizontal. If the tension in the rope is 200. N and the box accelerates to the right at 2.50 m/s2, what is the μ between the box and the floor?

Teacher Solution: CER Method

Claim $\mu \approx 0.21$.

Evidence $T_x = 160, T_y = 120$. $F_N = 400 - 120 = 280$. $f = 160 - 100 = 60$. $\mu = 60 / 280$.

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Additional Questions Video Instruction

1. How much tension must a steel cable withstand if it is used to accelerate a 1050 kilogram car horizontally at $1.20 \text{ m/s}^2$? Ignore friction.

2. What average force is required to stop a 1100 kilogram car in 8.0 seconds if it is traveling at 90. km/hr?

3. How much tension must a steel cable withstand if it is used to accelerate a 1200. kilogram crate vertically upwards at $0.80 \text{ m/s}^2$? Ignore friction.

4. A 10.-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket?

Teacher Solution: CER Method

Claim $a = 3.7 \text{ m/s}^2$ down.

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F = ma / Dragging a Box / Finding T

Find all Unknown Tensions Instructional Video

Horizontal Chains

Instructional Video

Vertical Tensions

Instructional Video 5 & 8

Pulley Combo Tension
Hanging Weight System

Pre - Class Questions

1. General Kinetic Friction Equation = f =

2. Normal force sitting on a ramp (Perpendicular component)

$F_n = $

3. Parallel component of the weight force

Ramp Diagram Pre-Class

$F_{parallel} = mg$

4. Calculate acceleration of a frictionless block down a ramp.

Teacher Solution: CER Method

Claim $a = 10 \sin \theta$.

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Example Problem 1

1. A 900. kg box slides down from rest along a wet slide inclined at an angle of 25 degrees to the horizontal. The coefficient of friction between the box and the slide is 0.0500.

 
Teacher Solution: CER Method

Claim $a \approx 3.77 \text{ m/s}^2$.

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Example Problem 2

2. Rather than taking the stairs, David likes to slide down the banister in his house. David has a mass of 20. kg, the banister has an angle of 30 degrees relative to the horizontal, and the coefficient of kinetic friction between David and the banister is 0.20. David begins his motion from rest.

 

a) What is the normal force between the banister and David?

b) What is the force of friction impeding his motion?

c) What is the net force on David?

d) How large is the acceleration as David slides down?

Teacher Solution: CER Method

Claim $a \approx 3.3 \text{ m/s}^2$.

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