2.7 Kinetic and Static Friction — Worked Solutions
Work through Topic 2.7
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 box stays at rest under a 6 N pull; its maximum static friction is 10 N. What is the friction magnitude?
- 10 N
- 6 N
- Zero
Step 1. Stationarity requires the friction force to balance the 6 N pull, so f_s = 6 N opposite the pull.
Why this step? A resting box has zero horizontal acceleration; with only these horizontal forces, their magnitudes must match.
Step 2. Check 6 N ≤ 10 N, so static friction can supply the required value.
Why this step? The stated maximum is a capacity limit, not the force that automatically acts whenever contact is static.
Step 3. Choose 6 N, not 10 N or zero.
Why this step? A 10 N force would leave an unbalanced resultant, and zero would leave the pull unbalanced; neither fits the stationary box.
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Question 2
A box rides without slipping on a belt that accelerates right. Which description can be correct?
- Static friction on the box points right
- Kinetic friction must point left
- No friction can act because there is no slip
Step 1. No slipping means the box is at rest relative to the contacting belt surface, so the friction is static.
Why this step? The classification uses relative contact motion rather than the box’s velocity through the room.
Step 2. To accelerate with the rightward-accelerating belt, the box needs a rightward net horizontal force.
Why this step? Without that forward force, the belt would slide ahead underneath the box.
Step 3. Static friction on the box can therefore point right.
Why this step? It opposes the box’s tendency to lag relative to the belt, so friction need not oppose the box’s ground velocity.
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Question 3
In the ideal kinetic-friction model, what sets its magnitude?
- μ_k times normal force
- μ_k times weight in every situation
- Contact area alone
Step 1. For dry sliding contact in this ideal model, use f_k = μ_kN.
Why this step? The coefficient describes the surface pair, and the perpendicular contact loading supplies the force scale.
Step 2. Determine N from the perpendicular force equation before substituting.
Why this step? An incline or an angled push can make N differ from mg; substituting weight by habit would use the wrong contact load.
Step 3. Choose μ_k times normal force.
Why this step? The model does not use apparent contact area alone, and μ_kmg is only a special case when N = mg.
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Question 4
A 2 kg block has μ_s = 0.40 and μ_k = 0.25 on a level floor. Determine its friction and acceleration under a 5 N pull, then under a 10 N pull starting from rest.
Step 1. On the level floor, vertical balance gives N = mg = 2·10 = 20 N.
Why this step? The pulls are horizontal and no vertical acceleration is stated, so weight is the only downward load.
Step 2. The static maximum is μ_sN = 0.40·20 = 8 N.
Why this step? This threshold decides whether a box initially at rest can remain in the no-slip regime.
Step 3. For a 5 N pull, static friction is 5 N and acceleration is zero.
Why this step? The required opposing force is below 8 N, so static friction can adjust to balance the pull exactly.
Step 4. For a 10 N pull from rest, the 8 N static capacity is exceeded. Once sliding, f_k = 0.25·20 = 5 N.
Why this step? The onset of slip changes the appropriate coefficient; the static maximum cannot be used as the continuing sliding force.
Step 5. Then a = (10 − 5)/2 = 2.5 m/s² in the pull direction.
Why this step? The kinetic force opposes motion, leaving a positive 5 N resultant acting on the 2 kg mass.
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Question 5
Derive the largest angle at which a block can remain at rest on an incline. Explain why changing mass alone doesn’t change that angle in this model.
Step 1. For a block at rest, N = mg cos θ and required uphill friction is mg sin θ.
Why this step? The incline constraint removes perpendicular acceleration, while the downhill gravity component would cause slip without an uphill contact force.
Step 2. No slip requires mg sin θ ≤ μ_smg cos θ.
Why this step? Static friction can supply the required value only up to its normal-force-dependent maximum.
Step 3. At impending slip, equality gives tan θ_max = μ_s, or θ_max = arctan μ_s.
Why this step? Dividing by positive mg cos θ converts the two component factors into a tangent; equality identifies the largest allowable angle.
Step 4. The mass cancels, so changing mass alone does not change θ_max in this model.
Why this step? The extra mass increases both the downhill demand and friction capacity in the same proportion, assuming the surface coefficient remains unchanged.
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Question 6
Sketch friction magnitude versus increasing horizontal pull for the box in Example 2.7.1, first at rest and then after slip. Mark the 20 N maximum and 12 N kinetic value.
Step 1. Use the example’s 4 kg box, N = 40 N, μ_s = 0.50, and μ_k = 0.30.
Why this step? The given graph must refer to this box’s fixed contact conditions, not an unspecified surface or changing vertical load.
Step 2. While the box remains at rest, draw f_s equal to pull magnitude up to f_s,max = 20 N.
Why this step? Zero acceleration requires matching horizontal forces, and the 0.50·40 limit is the point where that adjustment stops being possible.
Step 3. After slip begins, draw the kinetic branch at f_k = 12 N.
Why this step? Sliding uses μ_kN = 0.30·40, which is lower than the preceding static maximum even as the applied pull continues increasing.
Step 4. Label the first branch static and the second kinetic, with pull horizontally and friction magnitude vertically.
Why this step? The branch change is a contact-motion change; the graph shows friction magnitude, not the net force or signed acceleration.