Physics 201 Forces Exam Problem Example

John, a 19-year-old engineeringstudent, stared at the inclined plane diagram projected onto the lecture hall screen. Professor Chen's voice echoed, "This is a classic Physics 201 forces exam problem. You'll need to apply Newton's laws meticulously." The problem was:

Problem Statement: A 5.0 kg block rests on a frictionless inclined plane angled at 30 degrees to the horizontal. A constant horizontal force F is applied to the block, pushing it directly up the incline. The block accelerates up the incline at 2.0 m/s². Calculate the magnitude of force F.

John took a deep breath. This problem required synthesizing several core concepts: resolving forces on an incline, applying Newton's Second Law, and interpreting acceleration direction. He knew breaking it down systematically was key.

Step 1: Understand the System and Forces

• Mass (m): 5.0 kg
• Incline Angle (θ): 30°
• Acceleration (a): 2.0 m/s² up the incline
• Forces Acting on the Block:
  • Weight (mg): Acts vertically downward. Magnitude: m * g = 5.0 kg * 9.8 m/s² = 49 N.
  • Normal Force (N): Perpendicular to the incline surface.
  • Applied Force (F): Horizontal, pushing the block up the incline.
  • (Friction is absent, as stated - frictionless incline).

Step 2: Resolve Forces into Components The critical step is resolving forces parallel and perpendicular to the incline. The weight vector is the primary force needing resolution.

• Weight Components:
  • Parallel to incline: mg * sin(θ) = 49 N * sin(30°) = 49 N * 0.5 = 24.5 N down the incline.
  • Perpendicular to incline: mg * cos(θ) = 49 N * cos(30°) = 49 N * (√3/2) ≈ 49 N * 0.866 = 42.4 N into the incline.

Step 3: Apply Newton's Second Law (ΣF = ma) We focus on the parallel direction to the incline, as that's where the acceleration occurs. Define the positive direction up the incline.

Forces acting up the incline:

1. The applied force F (positive).
2. The component of weight opposing the motion (negative): -mg sinθ = -24.5 N.

Forces acting down the incline:

1. The component of weight acting down the incline: +mg sinθ = +24.5 N (negative in our up-incline positive system).

The net force acting up the incline is the sum of these forces. Newton's Second Law states this net force equals mass times acceleration in that direction:

ΣF_parallel (up incline) = m * a

Substituting the known values:

F + (-mg sinθ) = m * a

F - 24.5 N = (5.0 kg) * (2.0 m/s²)

F - 24.5 N = 10.0 kg·m/s²

F = 10.0 N + 24.5 N

Step 4: Calculate the Applied Force F = 34.5 N

Step 5: Verify Direction and Magnitude

• The force F is horizontal and pushes up the incline. Our calculation gives a positive value (34.5 N) in the direction we defined as positive (up the incline), confirming it's the correct direction.
• Magnitude: 34.5 Newtons.

Scientific Explanation: Why This Works This problem exemplifies the application of Newtonian mechanics to constrained motion on an inclined plane. The key insight is recognizing that the horizontal force F is not parallel to the incline. Its direction must be resolved into components parallel and perpendicular to the plane to accurately determine the net force causing acceleration along the incline.

The component of weight parallel to the incline (mg sinθ) always acts down the incline due to gravity. The horizontal force F, when resolved, has a component up the incline. The net force in the parallel direction determines the acceleration via ΣF = ma. The frictionless condition simplifies the analysis, eliminating the need to account for friction forces.

Understanding the geometry of the incline is crucial. The angle θ relates the vertical and horizontal components of forces. Trigonometric functions (sin, cos) are indispensable tools for resolving vectors into these components. Newton's Second Law provides the fundamental relationship linking force, mass, and acceleration, which is the cornerstone of analyzing dynamics problems like this.

FAQ

• Q: Why do we resolve forces parallel and perpendicular to the incline?
  • A: Because the block's acceleration is only along the incline. Forces perpendicular to the incline (like the normal force) cancel out in the direction of motion and are only relevant for determining the normal force itself. Resolving parallel allows us to directly apply ΣF = ma along the direction of acceleration.
• Q: What if the incline wasn't frictionless?
  • A: We would need to include the friction force. The friction force would act opposite the direction of motion (up the incline in this case) and its magnitude would be μ * N, where μ is the coefficient of friction and N is the normal force. Calculating N would require resolving the perpendicular components again. This adds complexity but follows the same systematic approach.
• Q: How do I draw the free-body diagram correctly?
  • A: Draw the block. Draw vectors representing each force acting on it. The weight vector is straight down. The normal force is perpendicular to the surface, pointing outward. The applied force is horizontal. Label all magnitudes and directions clearly. Ensure the diagram accurately reflects the geometry of the incline and the directions of the forces.
• Q: Why is the acceleration given as 2.0 m/s² up the incline?
  • A: This tells us the block is speeding up as it moves up the incline. The direction is crucial for setting the sign convention (+ up the incline)

The sign convention chosen for the parallel direction therefore sets the algebraic value of the net force equal to (ma) with a positive sign when the block accelerates upward. By writing the equation of motion along the incline we can solve for the unknown applied force:

[ \sum F_{\parallel}=F\cos\theta - mg\sin\theta = ma ]

where (F\cos\theta) is the component of the horizontal push that acts up the slope, (mg\sin\theta) is the weight component pulling the block down, and (ma) is the product of the block’s mass and its measured acceleration (2.0 m s⁻²). Rearranging gives

[ F = \frac{ma + mg\sin\theta}{\cos\theta} ]

Substituting the numerical values—(m = 5.0;\text{kg}), (a = 2.0;\text{m s}^{-2}), (g = 9.8;\text{m s}^{-2}), and an assumed incline angle (e.g., (\theta = 30^{\circ}))—yields a concrete magnitude for (F). The calculation illustrates how the resolved components interact: the horizontal push must not only overcome the downslope component of gravity but also provide the extra force required to generate the observed upward acceleration.

A few practical take‑aways emerge from this analysis:

1. Component resolution is non‑negotiable. Whether a force is applied horizontally, vertically, or at any arbitrary angle, projecting it onto the direction of interest simplifies the application of Newton’s second law.
2. Directionality dictates sign. The chosen positive direction (up the incline in this case) determines whether terms are added or subtracted in the force balance.
3. Frictionless assumptions streamline the algebra. By neglecting kinetic or static friction, the only forces that appear in the parallel equation are those explicitly resolved, allowing a clear view of cause and effect.
4. Iterative verification. After obtaining (F), one can check consistency by recomputing the net parallel force and confirming that it indeed equals (ma). If the result is inconsistent, revisit the geometry or the assumed angle.

In summary, the problem demonstrates the systematic approach to dynamics on an inclined plane: isolate the body, draw an accurate free‑body diagram, resolve every force into components aligned with the chosen axes, apply (\sum F = ma) along the relevant axis, and finally interpret the algebraic outcome in physical terms. Mastery of this workflow equips students to tackle a wide range of motion‑related questions, from simple textbook exercises to more complex engineering scenarios involving multiple interacting forces.