Questions: Equations of Motion from Free Body Diagrams

Question 1

A block of mass m sits on a frictionless inclined plane at angle θ. Aligning the x-axis along the slope (positive pointing down the slope) and the y-axis perpendicular to it, what is the net force equation along the x-axis?

• A. ΣFₓ = mg — the full weight acts along the slope
• B. ΣFₓ = mg cos θ — the component of gravity perpendicular to the slope
• C. ΣFₓ = mg sin θ — the component of gravity along the slope
• D. ΣFₓ = N − mg — normal force minus total weight

Answer: (C)

When the x-axis is aligned along the slope, the gravitational force (pointing straight down) has two components: mg sin θ along the slope and mg cos θ perpendicular to the slope. Only mg sin θ appears in the x-equation: ΣFₓ = mg sin θ = maₓ, giving acceleration aₓ = g sin θ down the slope. The normal force N points perpendicular to the surface, so it appears only in the y-equation (where it balances mg cos θ). This is the payoff of smart axis choice: the y-equation immediately gives N = mg cos θ, and the x-equation gives the acceleration without solving a system.

Question 2

A car of mass m is accelerating horizontally on a flat road. What is the normal force from the road on the car?

• A. Greater than mg, because the engine exerts a downward force through the drivetrain
• B. Less than mg, because the car is moving and kinetic effects reduce the apparent weight
• C. Equal to mg, because vertical acceleration is zero: ΣFᵧ = N − mg = maᵧ = 0, so N = mg
• D. Cannot be determined without knowing the car's speed or acceleration

Answer: (C)

The normal force is determined by the perpendicular force equation, independent of the horizontal motion. Since the car accelerates horizontally but not vertically, aᵧ = 0. Applying Newton's second law in the vertical direction: N − mg = 0, so N = mg. The horizontal acceleration is irrelevant to the vertical force balance. This shows why N = mg only holds when perpendicular acceleration is zero — on an incline, N = mg cos θ < mg; in an accelerating elevator, N ≠ mg. The formula N = mg is a consequence of zero vertical acceleration, not a general law.

Question 3

The direction of the net force on an object typically matches the direction of the object's velocity.

• T. True
• F. False

Answer: False

Net force determines the direction of acceleration, not velocity. A ball thrown upward has upward velocity but downward net force (gravity). A car braking while moving forward has forward velocity and backward net force. A planet in circular orbit has velocity tangent to the orbit but net force (gravity) pointing toward the center — perpendicular to velocity. Net force and velocity can point in the same direction, opposite directions, or at any angle. Confusing the direction of force with the direction of motion is one of the most common conceptual errors in introductory mechanics.

Question 4

On an inclined plane at angle θ, the normal force equals mg cos θ rather than mg, because it only needs to balance the component of gravity perpendicular to the surface.

• T. True
• F. False

Answer: True

Applying Newton's second law perpendicular to the incline: the block doesn't accelerate into or away from the surface, so ΣF_perp = N − mg cos θ = 0, giving N = mg cos θ. This is less than mg because gravity's full weight does not push directly into the surface — only its perpendicular component does. On flat ground θ = 0°, cos 0 = 1, and N = mg is recovered as a special case. This is why assuming N = mg on an incline leads to errors in friction calculations and in analyzing the block's motion.

Question 5

Why is the choice of coordinate axes so important when applying Newton's second law to an inclined plane, and what advantage does aligning the x-axis along the slope provide?

Model answer: Newton's second law must be applied component-by-component, so the axes determine how forces are decomposed. With horizontal/vertical axes, both the normal force and gravity have components in each direction, and the equations for x and y are coupled — you must solve them together as a system. Aligning the x-axis along the slope decouples the problem: the normal force lies entirely along the y-axis (perpendicular to slope), and the y-equation immediately gives N = mg cos θ. Gravity's component along the slope (mg sin θ) appears only in the x-equation, giving acceleration directly without simultaneous equations. The general principle is to align one axis with the direction of acceleration — this isolates the dynamics in one equation and the constraint forces in the other.

Smart axis choice doesn't change the physics, but it can eliminate an entire equation from the problem. This is why the methodology step 'choose axes' is not optional — a poor choice multiplies the algebra needed.