2.2 Forces and Free-Body Diagrams — Worked Solutions
Work through Topic 2.2
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
A block sits on a frictionless incline. Which forces belong on its FBD?
- Gravity, normal force, and an extra downhill component
- Gravity and normal force
- Gravity and a force of motion
Step 1. List the block’s interactions: Earth exerts gravity, and the incline exerts a normal force.
Why this step? Frictionless contact excludes friction, and no rope, hand, or other contact force is stated.
Step 2. Draw mg vertically down and N perpendicular outward from the incline, both starting at the block’s dot.
Why this step? Force direction comes from its source interaction; choosing inclined axes does not rotate gravity away from vertical.
Step 3. Use mg sin θ and mg cos θ only when writing component equations.
Why this step? They are projections of the same gravitational force; drawing them as extra forces would count gravity more than once.
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Question 2
A rope pulls a stationary crate partly upward. Compared with mg, the normal force is usually…
- Smaller
- Exactly equal
- Larger regardless of angle
Step 1. For a stationary crate, set a_y = 0 with upward positive: N + T_y − mg = 0.
Why this step? The floor and rope both push or pull upward, while gravity points down; stationarity requires their signed sum to vanish.
Step 2. Rearrange N = mg − T_y. Since the rope pulls partly upward, T_y is positive and N is smaller than mg.
Why this step? The rope carries some of the support otherwise supplied by the floor; this conclusion assumes the crate remains in contact.
Step 3. If the calculated N becomes negative, the contact model must change rather than allowing the floor to pull down.
Why this step? An ordinary floor normal force can push away from the surface but cannot hold the crate to it after separation.
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Question 3
Two actual forces on one object point downward. How should they appear on an AP FBD?
- As one unlabeled combined arrow
- As side-by-side arrows from the dot
- As arrows attached end to end
Step 1. Keep two separately labeled straight arrows, each beginning at the object’s dot and pointing downward.
Why this step? The two forces have distinct sources, so merging them would hide which interactions belong in the force sum.
Step 2. Place the arrows side by side so both remain visible.
Why this step? Overlapping arrows can conceal one force, while head-to-tail arrows represent vector addition rather than the required individual-force FBD.
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Question 4
Draw an FBD for a book pushed horizontally against a vertical wall and held stationary. Identify the source and direction of every force.
Step 1. Draw Earth’s gravitational force down and the hand’s applied push horizontally toward the wall.
Why this step? These are the stated source interactions; neither motion nor a generic “holding force” is an additional interaction.
Step 2. Draw the wall’s normal force horizontally away from the wall.
Why this step? A normal force is perpendicular to the contact surface; a vertical wall therefore has a horizontal normal direction.
Step 3. Draw static friction from the wall upward.
Why this step? Without friction, gravity would make the stationary book slip downward relative to the wall, so static friction must oppose that tendency.
Step 4. Start all four actual force arrows at the book’s dot and label their sources.
Why this step? The selected object is the book; forces the book exerts on the hand or wall belong on those other objects’ diagrams.
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Question 5
A 3 kg block rests on a horizontal floor. A downward push of 8 N is added. Find N using g = 10 m/s² and explain why it differs from weight.
Step 1. Weight is mg = 3·10 = 30 N downward, and the added push is 8 N downward.
Why this step? The push is a second real interaction; it cannot be absorbed into the definition of the block’s gravitational weight.
Step 2. Stationarity gives N − 30 − 8 = 0, so N = 38 N upward.
Why this step? The floor must balance both downward forces to maintain zero vertical acceleration, not gravity alone.
Step 3. Weight remains 30 N while the contact normal force rises to 38 N.
Why this step? Changing an applied push does not change mass or local g; normal force adjusts to the contact constraint.
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Question 6
For a block sliding down a rough incline, choose useful axes, draw only actual forces, then write both component equations without assuming constant speed.
Step 1. Choose x downhill along the incline and y perpendicular outward.
Why this step? The straight-track constraint makes perpendicular acceleration zero, leaving the changing motion in the parallel direction.
Step 2. Draw gravity vertically down, normal force outward perpendicular, and kinetic friction uphill.
Why this step? The block slides downhill relative to the rough surface, so friction opposes that slip; the gravity arrow is not redrawn along the slope.
Step 3. Write mg sin θ − f_k = ma_x.
Why this step? The gravitational parallel component points along positive x, while friction points against it; a_x is not set to zero without a constant-speed condition.
Step 4. Write N − mg cos θ = 0, with f_k = μ_kN if that coefficient model applies.
Why this step? Contact on a straight incline prevents perpendicular acceleration, and the normal force sets the kinetic-friction magnitude.