Dynamics and Forces Review
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Dynamics: Understanding Forces and Motion
Definition: Dynamics is the branch of mechanics concerned with the motion of bodies under the action of forces.
Two Important Words:
Forces: All discussions and work in dynamics involve understanding the forces acting on a system.
Motion: Dynamics problems begin by considering the forces on a system and conclude by understanding the system's motion (or vice-versa).
This material focuses heavily on examples rather than conceptual instruction.
What is a Force?
Definition: A force is a push or a pull that, when applied to an object, is capable of changing that object's velocity (its direction of motion or its speed).
Effect: A force causes an object to accelerate.
Quantity Type: Force is a vector quantity, meaning both magnitude and direction are important.
This necessitates the use of vector math, particularly components.
Unit of Measurement: Measured in Newtons (N).
Newton Definition: A Newton is a unit representing kg \cdot m/s^2.
Named After: Sir Isaac Newton (1642 - ~1726), who developed the foundational laws of motion.
Common Types of Forces:
Gravity
Tension
Normal force
PFG (likely a typo, might refer to other specific forces not fully elaborated)
Friction
Contact Requirement: Not all forces require direct contact (e.g., gravity).
Free-Body Diagrams (FBD)
Purpose: FBDs are a visual tool to perform the vector math required to sum 2D forces.
Steps for Drawing an FBD:
Represent the object being studied as a single dot.
Draw all forces acting on that object as arrows originating from the dot, pointing in their appropriate directions.
If necessary, separate these forces into their x and y components. Combine all vectors along the same axis, and then resolve the final components.
Example 1: Pulling a box across a rough floor with a rope at 30^\circ to the horizontal.
Forces involved: Applied force (Fa), Gravitational force (Fg), Normal force (FN), and Frictional force (Ff).
Example 2: Person in an elevator traveling at constant speed.
Forces: Tension from the elevator cable (T) and Gravitational force (F_g).
Since speed is constant, acceleration is zero, thus net force is zero. So, T and Fg are equal and opposite (T = Fg).
Example 3: Moving a piano up stairs with a rope and ramp (neglecting friction).
Forces: Tension from rope (T), Gravitational force (Fg), and Normal force (FN).
Sum of Forces: Terminology and Application
Symbol for "Sum of": The Greek letter sigma, \Sigma, is used to denote the sum.
"Sum of forces" in math: Written as \Sigma \vec{F}.
Resultant Force: The sum of all forces is often called the "resultant force" (\vec{F}_R) because it's the result of adding all other forces.
Vector Sum: This is a vector sum, requiring the rules of vector math.
For 2D problems, this means breaking forces into components (e.g., x and y components).
Component Summation:
\Sigma F_x means "the sum of forces in the x-direction" or "the sum of x-components of forces" or "the resultant force in the x-direction."
Similarly for \Sigma F_y.
Combining Forces Example:
Given |\vec{F}B| = 129 \ N and |\vec{F}A| = 111 \ N in a diagram where \vec{F}R is the red vector. To have zero net force, \vec{F}R must perfectly counteract \vec{F}A and \vec{F}B. Assuming \vec{F}A is in the negative y direction and \vec{F}B is in the negative x direction, then the components of \vec{F}_R will be:
\Sigma Fx = F{Rx} - F{Bx} = 0 \implies F{Rx} = F_{Bx} = 129 \ N
\Sigma Fy = F{Ry} - F{Ay} = 0 \implies F{Ry} = F_{Ay} = 111 \ N
The magnitude of \vec{F}_R is calculated using the Pythagorean theorem:
FR = \sqrt{F{Rx}^2 + F_{Ry}^2} = \sqrt{(129 \ N)^2 + (111 \ N)^2} = 170 \ N
The angle \theta of \vec{F}_R relative to the x-axis is:
\theta = \tan^{-1} \left( \frac{F{Ry}}{F{Rx}} \right) = \tan^{-1} \left( \frac{111 \ N}{129 \ N} \right) = 40.7^\circ
Newton's Second Law
Statement: "An object with mass m that has a non-zero net/resultant force \vec{F}R acting on it will experience an acceleration \vec{a} such that \vec{a} = \frac{\vec{F}R}{m}."
Commonly Shortened: F = ma. However, remember it's the net force that causes acceleration.
Component Form:
ax = \frac{F{R,x}}{m} or ax = \frac{\Sigma Fx}{m}
ay = \frac{F{R,y}}{m} or ay = \frac{\Sigma Fy}{m}
Units Consistency: The equation \vec{F} = m\vec{a} dictates that force \vec{F} must have units of kg \cdot m/s^2, which is defined as a Newton (N).
Force of Gravity: For gravity, Newton's Second Law becomes:
\vec{F}_G = m\vec{g}
Where g = 9.8 \ m/s^2, always pointing downward.
Example: Skydiver Acceleration
Scenario: A skydiver of mass 67 \ kg experiences an upward air resistance force of 567 \ N soon after jumping.
Goal: Determine the acceleration (magnitude and direction) of the skydiver.
Setup: Let the positive y-direction be upward.
\Sigma Fy = may = F{air} - Fg
Calculation:
ay = \frac{F{air} - Fg}{m} = \frac{F{air} - mg}{m}
a_y = \frac{567 \ N - (67 \ kg)(9.8 \ m/s^2)}{67 \ kg}
a_y = \frac{567 \ N - 656.6 \ N}{67 \ kg} = \frac{-89.6 \ N}{67 \ kg} = -1.337 \ m/s^2
Rounding to two significant digits (based on problem context implicitly):
a_y \approx -1.3 \ m/s^2
Conclusion: The acceleration is 1.3 \ m/s^2 downward.
Conceptual Questions on Newton's Laws
Car rounding a curve at constant speed:
Is there a non-zero net force? Yes. Even if speed is constant, the direction of velocity is changing, which means there is acceleration (centripetal acceleration), and thus a net force.
Elevator accelerating downward:
The upward tension force exerted on the elevator is smaller than the downward force of gravity acting on the elevator.
Since the elevator accelerates downward, the net force is downward. This means the downward gravitational force must be greater than the upward tension.
Elevator accelerating downward (person inside):
The upward normal force exerted on the person is smaller than the downward force of gravity acting on the person.
Similar to the elevator itself, if the person accelerates downward, the net force on them is downward. This implies the normal force from the floor supporting them is less than their weight.
Newton's First Law
Order of Laws: The First Law is often discussed after the Second Law because understanding \vec{F} = m\vec{a} makes the First Law easier to interpret.
Statement: "If the resultant force acting on an object is zero, the object does not accelerate – in other words, it continues at a constant velocity (a constant speed without changing direction)."
Common Phrasing: "Objects in motion tend to stay in motion, objects at rest tend to stay at rest, until acted on by an external, unbalanced force."
Mathematical Implication: From \vec{a} = \frac{\vec{F}_R}{m}, if \Sigma \vec{F} = 0, then \vec{a} must also be 0.
Examples Applying Newton's Laws
Example 1: Crate dragged at constant velocity
Scenario: A crate of mass 2.17 \ kg is dragged across a floor at a constant velocity of 0.200 \ m/s to the right with a horizontal rope. The floor exerts a frictional force of 6.05 \ N.
Goal a) Tension in the rope: Since velocity is constant, acceleration is zero, so \Sigma F_x = 0.
\Sigma Fx = FT - Ff = 0 \implies FT = F_f = 6.05 \ N
The tension in the rope is 6.05 \ N.
Goal b) Normal force exerted on the crate: Since there's no vertical acceleration, \Sigma F_y = 0.
\Sigma Fy = FN - Fg = 0 \implies FN = F_g
F_N = mg = (2.17 \ kg)(9.8 \ m/s^2) = 21.266 \ N
Rounding to three significant digits: F_N = 21.3 \ N.
Example 2: Child's toy car pushed by broom handle
Scenario: A child's toy car and kid have a combined mass of 36.7 \ kg. A force of 72.7 \ N is applied along a broom handle, making an angle of 69.3^\circ with the vertical. There is a frictional force of 12.7 \ N. The child starts from rest.
Goal: How far will the car travel in 1.50 \ s?
Setup: Define +x to the right and +y upward. The force from the broom (F_b) has components:
F{bx} = Fb \sin(69.3^\circ) (since the angle is with the vertical, the x-component uses sine)
F{by} = Fb \cos(69.3^\circ)
Calculate acceleration (x-direction):
\Sigma Fx = max = F{bx} - Ff
ax = \frac{F{bx} - Ff}{m} = \frac{Fb \sin(69.3^\circ) - F_f}{m}
a_x = \frac{(72.7 \ N) \sin(69.3^\circ) - 12.7 \ N}{36.7 \ kg}
a_x = \frac{67.92 \ N - 12.7 \ N}{36.7 \ kg} = \frac{55.22 \ N}{36.7 \ kg} = 1.5046 \ m/s^2
Rounding to three significant digits: a_x = 1.50 \ m/s^2
Calculate distance (kinematics): The car starts from rest (v_{0x} = 0).
\Delta x = v{0x}t + \frac{1}{2}axt^2
\Delta x = (0)(1.50 \ s) + \frac{1}{2}(1.5046 \ m/s^2)(1.50 \ s)^2
\Delta x = \frac{1}{2}(1.5046 \ m/s^2)(2.25 \ s^2) = 1.692675 \ m
Rounding to three significant digits: \Delta x = 1.69 \ m
Motion on an Inclined Plane
Effect of Incline Angle (\theta): As the incline angle \theta gets steeper, the acceleration of the object increases (neglecting friction).
General Acceleration (neglecting friction):
a = g \sin \theta along the incline.
Special Cases:
When \theta = 0^\circ (flat ground), a = g \sin(0^\circ) = 0.
When \theta = 90^\circ (vertical, free fall), a = g \sin(90^\circ) = g.
How to Approach Inclined Plane Problems
Coordinate System Strategy:
Choose the x-axis to lie along the hill (usually downhill, unless movement is uphill). The y-axis will then be perpendicular to the hill.
This choice simplifies forces: most forces (like normal force, friction, tension through a rope parallel to the incline) will lie entirely along one axis, requiring no component breakdown for them.
Gravity: The force of gravity (\vec{F}_g) will always need to be broken into components relative to this rotated coordinate system:
F{gx} = Fg \sin \theta = mg \sin \theta (component along the incline)
F{gy} = Fg \cos \theta = mg \cos \theta (component perpendicular to the incline)
Note: The angle \theta of the incline is the same angle between \vec{F}_g and the negative y-axis in the rotated coordinate system.
Example: Sled down a hill
Scenario: A sled starts from rest down a 100 \ m long hill at 18^\circ to the horizontal. Combined mass of sled and rider is 71 \ kg. Neglect friction. Assume 2 significant digits.
Goal a) Magnitude of the normal force (F_N):
In the y-direction (perpendicular to the incline), there is no acceleration, so \Sigma F_y = 0.
FN - F{gy} = 0 \implies FN = F{gy}
F_N = mg \cos \theta = (71 \ kg)(9.8 \ m/s^2) \cos(18^\circ)
F_N = (695.8 \ N)(0.9510) = 661.7 \ N
Rounding to two significant digits: F_N = 660 \ N.
Goal b) Acceleration down the slope (a_x):
In the x-direction (along the incline), the net force is due to the x-component of gravity.
\Sigma Fx = max = F_{gx}
max = mg \sin \theta \implies ax = g \sin \theta
a_x = (9.8 \ m/s^2) \sin(18^\circ) = (9.8 \ m/s^2)(0.3090) = 3.0282 \ m/s^2
Rounding to two significant digits: a_x = 3.0 \ m/s^2.
Goal c) Sled's speed at the end of the run (v_x):
Use kinematics: v{x}^2 = v{0x}^2 + 2a_x \Delta x
Given: v{0x} = 0 (starts from rest), ax = 3.0282 \ m/s^2, \Delta x = 100 \ m.
v_{x}^2 = 0^2 + 2(3.0282 \ m/s^2)(100 \ m) = 605.64 \ (m/s)^2
v_x = \sqrt{605.64} \ m/s = 24.609 \ m/s
Rounding to two significant digits: v_x = 25 \ m/s.
Multiple Objects & Inclined Plane Example
Scenario: Block A (9.0 \ kg) is hanging vertically, connected by a rope over a pulley to Block B (15 \ kg) on an inclined plane at 33^\circ. Neglect friction on the incline. Consider the system's acceleration.
Goal: Determine the magnitude and direction of the acceleration of the 9.0 \ kg block (Block A).
Setup: Assume Block A accelerates downward and Block B accelerates up the incline. Let downward for A and up the incline for B be positive directions.
Free Body Diagram for Block A (hanging):
Forces: Tension (FT) upward, Gravity (F{gA}) downward.
\Sigma FA = mA a = F{gA} - FT
mA a = mA g - FT \implies FT = mA g - mA a
Free Body Diagram for Block B (on incline):
Forces: Tension (FT) up the incline, x-component of Gravity (F{gBx}) down the incline, Normal force (F_N) perpendicular to incline.
\Sigma Fx = mB a = FT - F{gBx}
mB a = FT - mB g \sin(33^\circ) \implies FT = mB a + mB g \sin(33^\circ)
Equating Tension and Solving for Acceleration (a): Since the tension is the same for both blocks (ideal rope and pulley):
mA g - mA a = mB a + mB g \sin(33^\circ)
Rearrange to solve for a:
mA g - mB g \sin(33^\circ) = mA a + mB a
g(mA - mB \sin(33^\circ)) = a(mA + mB)
a = \frac{g(mA - mB \sin(33^\circ))}{mA + mB}
Substitute values: mA = 9.0 \ kg, mB = 15 \ kg, g = 9.8 \ m/s^2, \sin(33^\circ) \approx 0.5446
a = \frac{(9.8 \ m/s^2)((9.0 \ kg) - (15 \ kg)(0.5446))}{(9.0 \ kg) + (15 \ kg)}
a = \frac{(9.8 \ m/s^2)(9.0 \ kg - 8.169 \ kg)}{24.0 \ kg}
a = \frac{(9.8 \ m/s^2)(0.831 \ kg)}{24.0 \ kg} = 0.3392 \ m/s^2
Conclusion: The acceleration is 0.34 \ m/s^2. Since we set the direction for Block A downward as positive and the result is positive, Block A is indeed accelerating downward at 0.34 \ m/s^2. Consequently, Block B accelerates up the incline at 0.34 \ m/s^2.
Conclusion and Importance of Practice
Key Concept: Dynamics problems essentially involve understanding various forces and then applying Newton's Second Law, \vec{F} = m\vec{a}, using vector (component) math.
Resource Recommendation: The textbook likely contains extended sections (e.g., 5.5 and 5.6) with numerous practice problems for the First and Second Laws.
Problem-Solving Strategy (p. 121 in text): The text provides a step-by-step strategy for 2D dynamics problems using components.
Creating Your Own Plans: You can develop similar structured plans for specific problem types, such as "things suspended by rope tension" problems.
Beyond Memorization: Simply memorizing formulas like \vec{F} = m\vec{a} is insufficient. Success in dynamics requires identifying patterns in problem types and applying principles strategically.
Key to Success: Consistent practice is essential to develop these problem-solving skills and intuition. This is where strict memorization will fail you. You need to see the patterns, which comes with practice.