Chapter 1 - Forces Notes

Forces

Definition of Force

• Forces (F) are defined as something that causes an object to:

  • Start moving

  • Stop moving

  • Speed up

  • Slow down

  • Change direction

• Formula: $F = ma$

  • Where:

    • $F$ represents force.

    • $m$ represents mass.

    • $a$ represents acceleration.

• Unit of Force:

  • $(kg)(m/s^2)$

  • Newton (N)

  • $1 N = 0.225 lb$

Practice Problem

• How much force can be generated by a 5 kg object accelerating at a rate of 20 m/s²?

  • $F = ma$

  • $F = (5 kg)(20 m/s^2)$

  • $F = 100 N$

Classifying Forces - External Forces

• External Forces:

  • Forces that act on an object by interaction with the environment.

  • Produce movement ($\Delta$ motion).

  • Types:

    • Contact Forces

    • Non-contact Forces (gravity, magnetic force, electric force)

Classifying Forces - Internal Forces

• Internal Forces:

  • Forces that act within an object.

  • Types:

    • Tensile Force (tension, pull)

    • Compressive Force (compression, push)

  • When tensile and compressive forces are greater than what the structure can hold, it results in structural failure.

Classifying Forces - Gravity

• Gravity (g):

  • Force acting towards the center of the attracting body (Earth).

  • Imposes an effective acceleration of $9.81 m/s^2$ (at sea level).

  • Gravitational acceleration = Acceleration due to gravity.

Classifying Forces - Weight

• Weight (w):

  • A special case of force where $a = g$

  • Formula: $W = mg$

  • Unit: N (Newton)

Weight Practice Problem

• At sea level, what is the weight of an object that has a mass of 10 kg?

  • $W = mg$

  • $W = (10 kg)(9.81 m/s^2)$

  • $W = 98.1 N$

Review of Equations

• Velocity: $v = \Delta p/ \Delta t = d/ \Delta t$

  • Unit: m/s

• Acceleration: $a = \Delta v/ \Delta t = (v<em>2 - v</em>1)/(t<em>2 - t</em>1)$

  • Unit: $m/s^2 = (m/s)/s$

• Acceleration due to gravity: $g = 9.81 m/s^2$

• Force: $F = ma$

  • Unit: N = $(kg)(m/s^2)$

• Weight: $W = mg$

  • Unit: N

Practice Problem - Weight Calculation

• A steel shot has a mass of 7.257 kg. What is its weight?

  • $W = mg$

  • $W = (7.257 kg)(9.81 m/s^2)$

  • $W = 71.19 N$

Practice Problem - Velocity and Distance

• If a man is running at 4.48 m/s, how long will it take him to run 10 km? (Convert km to m)

  • $v = d/ \Delta t$

  • $4.48 m/s = 10000 m / \Delta t$

  • $\Delta t = 10000 m / 4.48 m/s$

  • $\Delta t = 2232 s$

Practice Problem - Acceleration

• A cyclist begins to accelerate at a rate of $0.45 m/s^2$. After 8 seconds of acceleration, the cyclist's new velocity is $10.9 m/s$. What was the initial velocity?

  • $a = (v<em>f - v</em>i) / \Delta t$

  • $0.45 m/s^2 = (10.9 m/s - v_i) / 8 s$

  • $v<em>i = v</em>f - a \Delta t$

  • $v_i = 10.9 m/s - (0.45 m/s^2)(8 s)$

  • $v_i = 7.3 m/s$

Practice Problem - Free Fall

• An object is dropped from a 10 m height. If there is no air resistance, what would be its velocity after falling for 1.1 s?

  • $v = g*t$

  • $v=9.81 m/s^2 * 1.1 s$

  • $v \approx 10.791 m/s$

Vectors

• Vectors represent dynamic action and have:

  • Magnitude (number)

  • Direction (orientation)

• Example: "35 mph due east" is a vector, while "35 mph" is not.

Vector Representation

• Vectors are represented by arrows:

  • Size indicates magnitude.

  • Angle indicates direction.

Examples of Vectors

• A velocity of 20 m/s acting due north.

• A force of 60 N acting horizontally.

• A weight of 300 N acting downwards.

Multiple Vectors

• Relative size of arrows indicates relative magnitude (e.g., 400 m/s vector is twice as long as a 200 m/s vector).

Vector Addition - Resultant

• Resultant:

  • The result of multiple vectors acting at the same time.

  • Addition of two or more forces.

Vector Addition - Co-linear Vectors (Same Direction)

• For co-linear vectors in the same direction, add the magnitudes.

Vector Addition - Co-linear Vectors (Opposite Directions)

• For co-linear vectors in opposite directions, subtract the magnitudes.

Vector Addition - Non-Co-linear Vectors

• Place vectors together using the tip-to-tail method.

• Use trigonometry to find the resultant vector.

Addition of Non-Co-linear Vectors - Law of Cosines

• To find the magnitude of the resultant vector, use the Law of Cosines.

• $C^2 = A^2 + B^2 - 2AB \cos(\gamma)$

  • Where C is the resultant vector, A and B are the magnitudes of the other two vectors, and $\gamma$ is the angle between vectors A and B.

Addition of Non-Co-linear Vectors - Law of Sines

• Use the Law of Sines to find the direction of resultant vector.

• $\frac{\sin(a)}{A} = \frac{\sin(b)}{B} = \frac{\sin(c)}{C}$

Vector Separation

• A single vector (force) acting at an angle can be separated into two perpendicular vectors (component vectors).

  • Vertical Component

  • Horizontal Component

• Component vectors represent the effect of the original vector in different directions.

Vector Separation - Rectangle Method

• Consider the original vector as the diagonal of a rectangle, with the component vectors as the sides.

Vector Separation - Trigonometry

• Use sine and cosine functions to calculate the magnitudes of component vectors.

  • $\sin(\theta) = \frac{V_v}{c}$

  • $\cos(\theta) = \frac{V_H}{c}$

  • $V_H = c \cos(\theta)$

  • $V_v = c \sin(\theta)$

Vector Separation - Example

• Given a vector $c = 50 N$ at an angle of 60°:

  • $V_H = 50 N * \cos(60°) = 25 N$

  • $V_v = 50 N * \sin(60°) \approx 43.5 N$

F Separation

• Contact forces can be separated into two components:

  • Perpendicular to the surface (vertical)

  • Parallel to the surface (horizontal)

Forces Affecting Movement - Friction

• Friction:

  • A resisting force.

  • Normal ($\perp$) to the reaction force.

  • Types:

    • Static Friction: Friction during non-movement. As sliding force increases, static friction increases equally, up to a maximum static friction ($F_m$).

    • Dynamic Friction: Friction during movement (kinetic friction, $F_k$).

      • $F_k$ is constant once an object starts to slide.

      • $F<em>k$ is always less than $F</em>m$.

Forces Affecting Movement - Rolling Friction

• Rolling friction resists the movement of an object rolling on a given surface.

Factors Affecting Friction

• Coefficient of Friction ($\mu$):

  • Relative difficulty of sliding.

  • Depends on roughness of the surfaces and molecular interaction.

• Normal Reaction Force (R):

  • Equal in magnitude and opposite in direction to the object's weight.

Friction Calculations

• Formula: $f = \mu R$

Friction Problems - Example 1

• The coefficient of static friction between the sole of Ken's shoes and the basketball court floor is 0.67. If Ken exerts a normal contact force of 1400 N when he pushes off the floor to run down the court, how large is the friction force exerted by Ken's shoes on the floor?

  • $f = \mu R$

  • $f = 0.67 * 1400 N$

  • $f = 938 N$

Friction Problems - Example 2

• Billy is trying to slide an 80-kg box of equipment across the floor. The coefficient of static friction between the box and the floor is 0.55. If Billy pushes only sideways (horizontally) against the box, how much force must he push with to initiate movement of the box?

  • Convert mass to weight: $W = mg = 80 kg * 9.81 m/s^2 = 784.8 N$

  • $f = \mu R$

  • $f = 0.55 * 784.8 N$

  • $f = 431.6 N$

Static Equilibrium

• When an object is at rest, the external forces acting on the object are in equilibrium.

  • Net force is zero.

  • All the forces acting on the object are cancelled out.

  • $\Sigma F = 0$

Static Equilibrium Problem

• Katie is exerting a 400 N upward force on a 700 N barbell that is resting on the floor. The barbell does not move. How large is the normal reaction force exerted by the floor on the barbell?

  • The net force is downward, so the normal reaction force must balance the difference between the weight of the barbell and Katie's upward force.

  • $700 N - 400 N = 300 N$

  • The normal reaction force exerted by the floor on the barbell is 300 N.

Practice Problems Part 1

Carlos assists Paul in lifting a 980 N barbell, Carlos exerts a 50 N upward force, and Paul exerts a 970 N force. What is the net vertical force exerted on the barbell?

• Net vertical force = $970 N + 50 N - 980 N = 40 N$

Based on the image above, find Beta

Given that $C^2= A^2 +B^2-2AB (Cos\gamma)$

• $\gamma = 180 - 120 = 60$

• $C^2 = 100^2 + 40^2 - 2(100)(40)cos(60)$

• $C^2 = 11,600 - 4000$

• $C = \sqrt{7600} \approx 87.18N$
  Now, using the Law of Sines: $\frac{sin B}{B} = \frac{sin \gamma}{C}$

• $sin B = \frac{B sin \gamma}{C}$

• $\frac{40 sin 60}{87.18} = 0.397$

• B = arcsin(0.397) = 23.39 degrees

Skateboarder problem

• $v<em>i = v</em>f - a \Delta t$

• $V_i= 5.2 m/s -( 0.20 m/s^2)(12 s) = 2.8 m/s$

Running problem

Given $v = \frac{d}{\Delta t}$

• $\frac{8000 m}{4.18 m/s} = 1913.8 s$

Assignment #2

Force needed to accelerate an object Problem

With a $6 kg$ object at a rate of $7m/s^2$

• $F=ma$

• $F = 6kg*7m/s^2$

• $F=42N$

External Forces

• External forces produce movement because they interact w/ the object from outside it's own system, causing changes in its States of motion. These forces can include gravity, friction, of applied forces like a push or pull. When an external force acts on an object, it can accelerate, decelerate, OR change direction, resulting in movement This it is described by Newton's law of motion, particularly the second law, which states that force equals mass times acceleration (F=m.a)

Internal Forces

• Internal forces act w/in an object and are important for understanding nature and causes of injury.

• These forces are improtant for understanding the nature and causes, as they relate to the stresses and strains that occur within tissues, muscles and bones during movement & impact. By studying internal forces, we can better comprehend how injuries happen and develop strategies to prevent them.

Non-Contact Force Examples

1. Gravitational Force: The force of attraction between two masses, such as the Earth & a falling object

2. Magnetic Force: The force exerted by a magnet on a metal object of another magnet, which can act over a distance without direct Contact.

Contact Force Examples

1. Frictional Force

2. Tension force

Gravity

• Gravity's a force acting toward the center of the earth. It imposes an effective acceleration value Of 9.8m/s² at sea level.

Dumbbell Weight Problem

Given a mass of 20kg for a dumbell what is it's weight
*Weight (W) = mass (m) x gravitational (g) acceleration
*m = 20kg
*g=9.8m/s²
*W= 20 kg x 9.8m/s²
*W = 196 N

Body Weight Incorrect Statement

Kilograms measure mass and not weight. A correct statement is "My body mass is 90kg" OR "My body weight is appx 882N"

Resultant Magnitude and Angle Calculation

Resultant magnitude calculation using: $C^2 = A^2 + B^2 -2<em>A</em>B*Cos(\gamma)$
*A = 80N
*B = 60
*$\gamma = 120$
$C^2 = 80^2 + 60^2 - 2</em>80<em>60</em>Cos(120)$
*Cos(120)= -.05
*$C^2 = 6400 + 3600 - 2<em>80</em>60*(-0.5)$
*$C^2 = 14800$
*$C = \sqrt{14800} \approx 121.65N$
Now find the angle:
$\frac{Sin(\alpha)}{A} = \frac{Sin(\beta)}{B} = \frac{Sin(\gamma)}{c}$
*Sin(120) = √3/2
*$\frac{121.65}{sin(120)} \frac{30 \sqrt{3}}{121.65} \approx 0.257$
Inverse B=arcsin(0.257)
*b = 14.9
Resultant Force will be appx 121.65N at an angle of 105. 1° from the VH

Horizontal and vertical components of a vector

Horizontal and vertical components of a 65 N vector that acts at an angle of 50 degrees above the horizontal.
*Fx = F * Cos(Θ)
*Fy = F * Sin(Θ)
Fx=65NCOS (50)
Fx=65N0.643
*Fx=41.8N
*Fy=65 N * Sin (50)
*Fy=65N * 0.766
*Fy=49.79 ≈49.8N

Fridge Sliding Problem

*Convert 200 lbs to kg
*1kg=2.2lb
*200/2.2=90.9
*f=µR
*90.9 * 9.81 =891.8
f=0.50*891.8
*f= 445.9N so at least 445.9 N is needed

Find side c of a triangle

Given a triangle find side c:
*a=8cm
*b=5cm
*$\gamma = 60$
*$C^2 = A^2 + B^2 -2AB*Cos(\gamma)$
*$C^2 = 8^2 + 5^2 -2(8)(5)*Cos(60)$
*$C^2 = 64 + 25 - 40$
*$C^2 = 49$
*C=√49
*C=7cm