Exam Review: Forces, Friction, Newton's Laws, and Introduction to Work

Homework Set 10: Connected Objects (No Friction)

• Problem Description: Two ideally connected objects on an inclined plane. The string and pulley are ideal (massless, inextensible).
• Assumptions: No friction (coefficient of static friction $\mu<em>s = 0$, kinetic friction $\mu</em>k = 0$).
• Ideal System Implications:
  • Ideal String: Massless, inextensible. This means the acceleration ($a$) is the same for both connected objects.
  • Ideal Pulley: Massless, frictionless. This means the tension ($T$) is the same on both sides of the pulley.
• Newton's Third Law Connection: Even without an ideal pulley, tensions would be equal and opposite due to Newton's Third Law (if considering string tension as an internal force between objects).
• Free Body Diagrams (FBDs):
  • Hanging Mass ($m<em>1$): Downward force of weight ($m</em>1g$), upward force of tension ($T$).
  • Mass on Plane ($m<em>2$): Downward force of weight ($m</em>2g$), normal force ($N$) perpendicular to the plane, tension ($T$) parallel to the plane.
• Breaking Up Gravity: For the mass on the inclined plane ($m_2$):
  • Perpendicular to the plane: $m_2g \cos \theta$
  • Parallel to the plane: $m_2g \sin \theta$
  • Common Mistake: Reversing sine and cosine for these components.
• Equations of Motion (Assuming $m<em>1$ moves up and $m</em>2$ moves down the incline - based on class convention):
  • For Hanging Mass ($m<em>1$): $T - m</em>1g = m_1a$
  • For Mass on Plane ($m_2$):
    • Perpendicular to plane (y-direction): $N - m_2g \cos \theta = 0$ (since the object stays on the plane).
    • Parallel to plane (x-direction): $m<em>2g \sin \theta - T = m</em>2a$ (assuming $m<em>2$ moves down the incline, so $m</em>2g \sin \theta$ is the driving force).
• Solving for Acceleration ($a$): By adding the two equations of motion (for $m<em>1$ and $m</em>2$), tension cancels out (Newton's Third Law at play, as it's an internal force for the system).
  $a = \frac{m<em>2 \sin \theta - m</em>1}{m<em>1 + m</em>2} g$
• Solving for Tension ($T$): Substitute $a$ back into one of the FBD equations (e.g., $T = m<em>1(a+g)$). $T = m</em>1 g \left( \frac{m<em>2 \sin \theta - m</em>1}{m<em>1 + m</em>2} + 1 \right) = \frac{m<em>1 m</em>2 g (\sin \theta + 1)}{m<em>1 + m</em>2}$
• Algebraic Precision: Emphasized the importance of meticulous algebra to avoid errors (e.g., common mistakes in bringing terms under a common denominator).
• Notation: Using $m<em>1, m</em>2$ is preferred over $M, m$ to prevent confusion, as they can sometimes look similar.
• Intuition Training: Special Cases:
  • Purpose: To verify formula correctness and build physical intuition.
  • Dimensional Analysis: Always check if the units on both sides of the equation match (e.g., force = mass * acceleration).
  • Case 1: Angle $\theta = \frac{\pi}{2}$ (Vertical Plane):
    • This means both masses are effectively hanging. If $m<em>1 = m</em>2$, the system is balanced.
    • Acceleration ($a$) becomes $0$ when $m<em>1 = m</em>2$. This makes sense for a balanced system.
    • Tension ($T$) becomes $mg$ when $m<em>1 = m</em>2$. This also makes sense as it's supporting the full weight of either mass in a balanced state.
  • Case 2: Angle $\theta = 0$ (Horizontal Plane):
    • The mass $m<em>2$ is on a flat horizontal surface. If $m</em>1 = m_2$ and no friction, the system will accelerate.
    • Acceleration ($a$) becomes $\frac{-m<em>1}{m</em>1+m<em>2} g$ (if $m</em>2$ is on the horizontal plane and $m<em>1$ is hanging). If $m</em>1 = m<em>2$, then $a = -\frac{1}{2}g$. The negative indicates motion opposite to the initial assumed positive direction for $m</em>1$, or that $m<em>1$ falls, dragging $m</em>2$.
    • Tension ($T$) becomes $\frac{m<em>1 m</em>2}{m<em>1 + m</em>2} g$. If $m<em>1 = m</em>2$, then $T = \frac{1}{2}mg$. The tension is less than $mg$ because the hanging mass is accelerating downwards, meaning the string does not have to support its full weight.

Homework: Snowboarder Problem (Kinetic Friction)

• Problem Description: A snowboarder landing on a steep slope, slowing down due to kinetic friction. Data: initial speed, final speed, slope angle.
• Assumptions: Uniform kinetic friction ($\mu_k \text{ constant}$), no air resistance.
• Forces on Snowboarder: Normal force ($N$), weight ($mg$), kinetic friction ($f_k$).
• Breaking Up Gravity: Same as before, $mg \cos \theta$ perpendicular and $mg \sin \theta$ parallel to the slope.
• Equations of Motion (assuming positive direction down the slope):
  • Perpendicular to plane: $N - mg \cos \theta = 0 \implies N = mg \cos \theta$
  • Parallel to plane: $mg \sin \theta - f_k = ma$
• Kinetic Friction Force: $f<em>k = \mu</em>k N = \mu_k mg \cos \theta$
• Substituting Friction into Equation of Motion:
  $mg \sin \theta - \mu_k mg \cos \theta = ma$
• Solving for Acceleration ($a$): Mass ($m$) cancels out.
  $a = g(\sin \theta - \mu_k \cos \theta)$
• Determining Unknowns: Given initial speed ($v<em>0 = 45 \text{ mph} = 20.1 \text{ m/s}$), final speed ($v</em>f = 10 \text{ mph} = 4.5 \text{ m/s}$), slope angle ($\theta = 39^ ext{o}$), and distance along slope ($\Delta x$ - necessary for kinematics).
• Kinematics Application: Since $a$ is constant (as $g, \theta, \mu_k$ are constant), use constant acceleration kinematic equations.
  • One common equation: $v<em>f^2 = v</em>0^2 + 2a \Delta x$. (Note: A common error in the transcript was writing $v<em>x^2 - v</em>y^2$ for final minus initial velocity, and then dividing by $2a$, implying $(v<em>f^2 - v</em>0^2)/(2 \Delta x) = a$.)
• Calculation: Given the values, the acceleration was found to be approximately $-2.7 \text{ m/s}^2$ (negative because it's slowing down). Then, $\mu_k$ was calculated to be $1.52$.
• Unit Conversion: Emphasized converting English units (miles per hour) to MKS (meters per second) for consistency.

Quiz Review: Newton's Laws of Motion

• General Expectation: Be able to state the laws in words, provide examples, and ideally use both language and pictures.

Newton's First Law (Law of Inertia)

• Statement: An object at rest stays at rest, and an object in uniform motion (constant velocity in a straight line) stays in uniform motion, unless acted upon by an external (or unbalanced) force.
• Key Concepts:
  • Uniform Motion: Constant velocity, implying no acceleration (no change in magnitude or direction).
  • External Force: A force originating outside the system in question; internal forces do not change the system's overall motion.
• Example (Implied): Students staying in their chairs, rather than falling or flying, due to balanced forces.

Newton's Second Law ($F=ma$)

• Statement: The net force ($\Sigma F$ or $F_{net}$) acting on an object is equal to the product of its mass ($m$) and its acceleration ($a$).
  $\Sigma \vec{F} = m \vec{a}$
• Vector Equation: This law applies to each component (x, y, z) independently. The components are