2.7 Kinetic and Static Friction

1. A0.33
   A student who takes the friction force to be proportional to the contact area picks this: one third of the area gives one third of the force. The friction force depends on the normal force, which is unchanged.
2. B1.00 Correct
   On a level floor the normal force equals the brick's weight on either face, and the coefficient depends only on brick and concrete. In this model |F⃗_f,k| = μ_k F_N depends neither on the contact area nor on the speed, so the ratio is 1.00.
3. C3.00
   A student who reasons that a smaller face presses harder and so has more friction picks this. The force per unit area is three times larger, but the total normal force, which sets the friction, is the same.
4. D2.00
   A student who thinks kinetic friction grows with sliding speed picks this. μ_k F_N contains no speed, so doubling the speed leaves the kinetic friction force unchanged.

Working F_f,k = μ_k F_N with F_N = mg in both cases (level floor, no other vertical forces). Neither area nor speed enters, so F₂/F₁ = (μ_k mg)/(μ_k mg) = 1.00.

CED 2.7.A.1.ii · Read this in Fix

1. A42 N Correct
   Vertically, F_N + (100 N) sin 37° = mg, so F_N = 200 N − 60 N = 140 N. Then |F⃗_f,k| = μ_k F_N = 0.30 × 140 N = 42 N. (The horizontal pull, 80 N, exceeds this, so the box speeds up.)
2. B60 N
   A student who takes the normal force to equal the weight, 200 N, picks this: 0.30 × 200 N = 60 N. The rope's upward component, 60 N, supports part of the weight, so the floor pushes up with only 140 N.
3. C80 N
   A student who thinks kinetic friction balances the horizontal pull picks this: (100 N) cos 37° = 80 N. Kinetic friction is μ_k F_N = 42 N whatever the pull; the difference accelerates the box.
4. D36 N
   A student who uses the cosine component as the rope's vertical component picks this: F_N = 200 N − 80 N = 120 N and 0.30 × 120 N = 36 N. With the angle measured from the horizontal, the vertical component is (100 N) sin 37° = 60 N.

Working Vertical: F_N + F sin 37° − mg = 0 → F_N = (20 kg)(10 m/s²) − (100 N)(0.60) = 140 N. F_f,k = μ_k F_N = (0.30)(140 N) = 42 N.

CED 2.7.A.2 · Read this in Fix

1. AKinetic friction from the bed acts backward, since the crate is moving.
   A student who calls friction kinetic whenever the object moves relative to the ground picks this. The crate does not slide on the bed, so the friction is static, and it must point forward to speed the crate up.
2. BStatic friction from the bed acts forward, the way the crate accelerates. Correct
   The crate and the bed do not move relative to each other, so any friction between them is static. The crate speeds up forward, so the net horizontal force on it points forward, and the only horizontal force is static friction from the bed.
3. CNo horizontal force acts; the crate is carried along by the truck.
   A student who thinks an object can be carried along without a force picks this. The crate's velocity changes, so a net force acts on it; the bed exerts it through static friction.
4. DStatic friction from the bed acts backward, opposite to the crate's motion.
   A student who thinks friction always opposes an object's velocity picks this. A backward force would slow the crate down; static friction points forward here, in the direction of the crate's velocity.

CED 2.7.B.1 · Read this in Fix

1. A9.0 N
   A student who takes static friction to equal μ_s F_N in every case picks this. 9.0 N is the largest value static friction can take; against a 4.5 N push, 9.0 N of friction would accelerate the block backward, opposite to the push.
2. B6.0 N
   A student who uses μ_k F_N for any friction force picks this: 0.40 × 15 N = 6.0 N. The block does not slide, so the friction is static, and kinetic friction applies only once the surfaces slide.
3. C0.0 N
   A student who thinks friction acts only on moving objects picks this. With no friction the 4.5 N push would accelerate the block; it stays at rest because static friction balances the push.
4. D4.5 N Correct
   The maximum static friction is μ_s F_N = 0.60 × 15 N = 9.0 N. The 4.5 N push is less than that, so the block stays at rest and static friction takes the value that keeps the net force zero: 4.5 N, opposite to the push.

Working F_N = mg = (1.5 kg)(10 m/s²) = 15 N. F_f,s,max = μ_s F_N = (0.60)(15 N) = 9.0 N > 4.5 N, so no slipping; Newton's second law with a = 0 gives F_f,s = 4.5 N.

CED 2.7.B.2 · Read this in Fix

1. Aμ_s < μ_k, so a crate is harder to keep sliding than to start it
   A student who thinks sliding surfaces grip harder picks this. The everyday observation is the opposite: the push needed drops once a heavy box starts to move.
2. Bμ_s > μ_k, so a crate is easier to keep sliding than to start Correct
   For a given pair of surfaces μ_s is typically greater than μ_k. The push needed to start a crate sliding, μ_s F_N, is therefore larger than the push needed to keep it sliding at constant speed, μ_k F_N.
3. Cμ_s = μ_k, since both describe the same two surfaces in contact
   A student who thinks one pair of surfaces has one coefficient picks this. The two coefficients describe different situations, surfaces not slipping and surfaces sliding, and typically μ_s is the larger.
4. DIt depends on the crate's mass, as a heavier crate presses harder
   A student who thinks a coefficient changes with the load picks this. A heavier crate has a larger normal force, which changes both friction forces, but neither coefficient.

CED 2.7.B.3 · Read this in Fix

1. AKinetic friction, directed to the left
   A student who thinks friction opposes whatever motion is happening, here the belt's motion to the right, picks this. Friction on the block opposes the block's motion relative to the belt, which is to the left, so it points right.
2. BStatic friction, directed rightward
   A student who calls friction static whenever the object is at rest relative to the ground picks this. The block's surface slides over the belt's surface, so the friction is kinetic, even though the block is at rest.
3. CKinetic friction, acting to the right Correct
   The block's surface and the belt's surface move relative to each other, so the friction is kinetic. Relative to the belt, the block moves to the left, so kinetic friction on the block points to the right, opposite to that relative motion; the string's pull to the left balances it.
4. DNo friction, since the block is at rest
   A student who thinks friction acts only on moving objects picks this. The belt slides under the block, and if there were no friction, nothing would balance the string's pull and the block would not stay at rest.

CED 2.7.A.1 · Read this in Fix

1. Aμ_k mg, toward the left
   A student who sets friction opposite to each object's velocity relative to the floor picks this. The lower block moves left relative to the surface it rubs against, the upper block, so friction on it points right; it is the friction on the upper block that points left.
2. Bμ_k mg + μ_k Mg, rightward
   A student who takes the normal force between the blocks to be the weight of both picks this. The contact between the blocks supports only the upper block, so the normal force there is mg; the floor supports (m + M)g.
3. Czero, both move rightward
   A student who thinks objects moving in the same direction do not slide on each other picks this. The velocities differ by 3.0 m/s, so the surfaces slide and kinetic friction acts.
4. Dμ_k mg, toward the right Correct
   The surfaces slide on each other because the velocities differ. Relative to the upper block, the lower block moves to the left (2.0 m/s − 5.0 m/s = −3.0 m/s), so kinetic friction on it points to the right. The normal force between the blocks supports only the upper block, mg, so the friction has magnitude μ_k mg.

CED 2.7.A.1.i · Read this in Fix

1. A4.0 m/s²
   A student who puts friction up the slope, as for a block resting on a ramp, picks this: (10 m/s²)(0.60 − 0.20) = 4.0 m/s². The block slides up relative to the incline, so kinetic friction on it points down the slope.
2. B8.5 m/s²
   A student who uses F_N = mg on the incline picks this: (10 m/s²)(0.60 + 0.25) = 8.5 m/s². Perpendicular to the incline only the component mg cos 37° must be balanced, so F_N = 0.80mg.
3. C9.5 m/s²
   A student who swaps the sine and cosine components picks this: (10 m/s²)(0.80 + 0.25 × 0.60) = 9.5 m/s². For an incline angle θ, gravity's component along the slope is mg sin θ and the perpendicular one is mg cos θ.
4. D8.0 m/s² Correct
   While the block moves up, kinetic friction points down the slope, the same way as the gravitational component. F_N = mg cos 37°, so a = g sin 37° + μ_k g cos 37° = (10 m/s²)(0.60 + 0.25 × 0.80) = 8.0 m/s², directed down the slope.

Working Up the slope is the direction of motion; both forces along the slope point down it. F_N = mg cos 37°. ma = mg sin 37° + μ_k mg cos 37° → a = (10 m/s²)(0.60 + 0.25 × 0.80) = 8.0 m/s² (directed down the slope).

CED 2.7.A.2 · Read this in Fix

1. Aμ_P > μ_Q > μ_R
   A student who ranks the lines by the greatest friction force reached picks this: P ends at 20 N, Q at 15 N, R at 12 N. Those heights depend on how much mass was added; the coefficient is the slope of each line.
2. Bμ_Q > μ_P > μ_R
   A student who thinks heavier loads mean larger coefficients ranks the lines by the largest normal force used: Q (60 N), P (40 N), R (16 N). The coefficient is the ratio F_f,k/F_N, which is constant along each line.
3. Cμ_R > μ_P > μ_Q Correct
   Each line passes through the origin, so its slope F_f,k/F_N is the coefficient: R, 12 N/16 N = 0.75; P, 20 N/40 N = 0.50; Q, 15 N/60 N = 0.25. The coefficient depends on the pair of materials, so it differs between the blocks.
4. Dμ_P = μ_Q = μ_R
   A student who thinks the coefficient belongs to the track alone picks this, since all three slide on aluminum. The coefficient belongs to the pair of surfaces, and the three lines have different slopes.

Working Slopes: R 12/16 = 0.75; P 20/40 = 0.50; Q 15/60 = 0.25. So μ_R > μ_P > μ_Q.

CED 2.7.A.2.i · Read this in Fix

1. Aμ_k(mg cos θ + F sin θ) Correct
   Perpendicular to the incline the block does not accelerate. Gravity has a component mg cos θ into the surface, and the horizontal push, which makes the angle θ with the incline's surface, has a component F sin θ into the surface. So F_N = mg cos θ + F sin θ, and the kinetic friction is μ_k times that.
2. Bμ_k(mg sin θ + F cos θ)
   A student who swaps the sine and cosine components picks this. Check the limit θ → 0: on a level floor the normal force must be mg and a horizontal push adds nothing to it, which the key gives and this option does not.
3. Cμ_k(mg cos θ + F cos θ)
   A student who uses 'magnitude times cos θ' for the perpendicular component of every force, as for gravity, picks this. The horizontal push makes the angle θ with the surface, so its perpendicular component is F sin θ.
4. Dμ_k(mg cos θ − F sin θ)
   A student who applies the pulled-sled result, in which an angled force reduces the normal force, picks this. Here the push's perpendicular component points into the incline, so it increases the normal force.

Working Axes along and perpendicular to the incline. Perpendicular: F_N − mg cos θ − F sin θ = 0 (the horizontal force makes angle θ with the incline surface, so its perpendicular component is F sin θ). F_f,k = μ_k F_N = μ_k(mg cos θ + F sin θ). Check θ → 0: F_N → mg, as for a level floor with a horizontal push.

CED 2.7.A.2.ii · Read this in Fix

1. Amg(sin θ + μ_s cos θ) Correct
   Static friction adjusts as F grows: it points up the slope while F < mg sin θ, is zero when F = mg sin θ, and then points down the slope, opposing the block's tendency to slide up, until it reaches its maximum μ_s mg cos θ. Sliding up begins when F = mg sin θ + μ_s mg cos θ.
2. Bmg sin θ − μ_s mg cos θ
   A student who keeps static friction pointing up the slope picks this; that expression is the smallest force that keeps the block from sliding down. When the block is about to slide up, static friction points down the slope, adding to gravity's component.
3. Cmg (cos θ + μ_s sin θ)
   A student who swaps the sine and cosine components picks this. For a level surface (θ → 0) the push needed must approach μ_s mg, which the key gives and this option does not.
4. Dmg sin θ + μ_s(mg)
   A student who takes the normal force on the incline to be mg picks this: F = mg sin θ + μ_s mg. Perpendicular to the incline only gravity's component mg cos θ is balanced, so F_N = mg cos θ and the maximum static friction is μ_s mg cos θ.

Working Along the incline (up positive), at the onset of sliding up, static friction is at its maximum and points down the slope: F − mg sin θ − μ_s F_N = 0 with F_N = mg cos θ, so F = mg(sin θ + μ_s cos θ). For F < mg sin θ the static friction points up the slope; it reverses direction as F passes mg sin θ.

CED 2.7.B.2 · Read this in Fix

1. A0.30
   A student who takes the friction during sliding to be the maximum static friction picks this: 45 N/150 N = 0.30. The level part of the graph is kinetic friction, which is smaller; the maximum static friction is the peak, 60 N.
2. B4.00
   A student who uses the mass in place of the normal force picks this: 60/15 = 4.00. The normal force is the weight, mg = 150 N, so μ_s = 60 N/150 N.
3. C0.40 Correct
   Static friction matches the push up to its maximum, 60 N, where the crate starts to slide. F_N = mg = 150 N, so μ_s = F_f,s,max/F_N = 60 N/150 N = 0.40.
4. D1.00
   A student who reads the slope of the rising part of the graph as the coefficient picks this. That slope is 1 because static friction equals the push; a coefficient compares the friction force with the normal force, not with the push.

Working F_f,s,max = 60 N (peak of the graph). F_N = mg = (15 kg)(10 m/s²) = 150 N. μ_s = 60 N/150 N = 0.40. (The level part, 45 N, gives μ_k = 0.30.)

CED 2.7.B.2.ii · Read this in Fix

1. A25 N
   A student who leaves out the friction the upper block exerts on the lower block picks this, writing F = Ma = (5.0 kg)(5.0 m/s²). By Newton's third law the upper block pulls back on the lower block with 15 N, so F = 25 N + 15 N.
2. B40 N Correct
   Not slipping means no relative motion, so the blocks share one acceleration. Only static friction accelerates the upper block, so a ≤ μ_s g = 5.0 m/s². For the two blocks as one system, F = (3.0 kg + 5.0 kg)(5.0 m/s²) = 40 N.
3. C15 N
   A student who sets the applied force equal to the maximum friction between the blocks picks this: (0.50)(30 N) = 15 N. That friction only has to accelerate the 3.0 kg block; F must accelerate both blocks.
4. D55 N
   A student who adds the friction between the blocks to the force needed to accelerate both picks this: 40 N + 15 N. That friction is internal to the two-block system, so it does not enter F = (m + M)a.

Working Upper block (3.0 kg): F_f,s = ma ≤ μ_s mg → a_max = μ_s g = (0.50)(10 m/s²) = 5.0 m/s². Both blocks (friction internal): F = (3.0 kg + 5.0 kg)(5.0 m/s²) = 40 N. Check, lower block: F − 15 N = (5.0 kg)(5.0 m/s²) → F = 40 N.

CED 2.7.B.2.i · Read this in Fix

1. Aμ_s g
   A student who thinks friction stays at its maximum static value, μ_s mg, once sliding begins picks this: (2μ_s mg − μ_s mg)/m. Once the crate slides the friction is kinetic, μ_k mg, which is typically smaller.
2. B2μ_s g
   A student who thinks friction disappears once the crate breaks free picks this: a = 2μ_s mg/m. Kinetic friction, μ_k mg, acts for as long as the crate slides.
3. Cg(2μ_s−μ_k) Correct
   The push, 2μ_s mg, exceeds the maximum static friction, μ_s mg, so the crate slides. Friction is then kinetic, μ_k mg, so the net force is (2μ_s − μ_k)mg and a = g(2μ_s − μ_k).
4. D2μ_s g−μ_k
   A student who replaces the normal force by the mass writes the kinetic friction as μ_k m and picks this: a = (2μ_s mg − μ_k m)/m. The normal force is mg, so the friction is μ_k mg and a = g(2μ_s − μ_k).

Working F_f,s,max = μ_s mg < 2μ_s mg, so the crate slides at once and friction is kinetic: F_net = 2μ_s mg − μ_k mg, a = F_net/m = g(2μ_s − μ_k).

CED 2.7.B.3 · Read this in Fix

1. A1.00
   A student who takes the normal force to be mg on any incline picks this. Only gravity's component perpendicular to the surface, mg cos θ, presses the crate into the incline, and it shrinks as the incline steepens.
2. B1.73
   A student who uses mg sin θ as the perpendicular component picks this: sin 60°/sin 30° = 1.73. As θ → 0 the normal force on a level surface must approach mg, which cos θ gives and sin θ does not.
3. C0.50
   A student who reasons that doubling the angle halves the perpendicular component picks this. cos θ is not inversely proportional to θ: cos 60°/cos 30° = 0.58.
4. D0.58 Correct
   F_f,s,max = μ_s F_N, and the normal force is mg cos θ in both cases (the rope is parallel to the surface, so it adds nothing perpendicular to it). The ratio is cos 60°/cos 30° = 0.50/0.87 ≈ 0.58.

Working F_N = mg cos θ (rope parallel to the surface). Ratio = μ_s mg cos 60°/(μ_s mg cos 30°) = 0.500/0.866 = 0.58.

CED 2.7.B.2.ii · Read this in Fix

1. A1.0μ_s mg
   A student who equates the largest applied force with the maximum static friction between the blocks picks this. The applied force must also accelerate the upper block: F − f = ma, so F exceeds the friction by ma.
2. B3.0μ_s mg
   A student who caps the common acceleration at μ_s g picks this: (3m)(μ_s g). That cap applies when the force is on the lower block and friction alone accelerates the upper block. Here friction alone accelerates the lower block, of mass 2m, so the cap is μ_s mg/(2m).
3. C1.5μ_s mg Correct
   Only friction from the upper block pushes the lower block, and it is at most μ_s mg, so the lower block's acceleration is at most μ_s mg/(2m) = 0.5μ_s g. Moving together, both blocks have this acceleration, so F_max = (3m)(0.5μ_s g) = 1.5μ_s mg.
4. D4.5μ_s mg
   A student who takes the normal force between the blocks to be the weight of both, 3mg, picks this: friction up to 3μ_s mg gives a_max = 1.5μ_s g and F = (3m)(1.5μ_s g). The contact between the blocks supports only the upper block, so the normal force there is mg.

Working Upper block: the normal force from the lower block is mg, so the static friction between the blocks is at most μ_s mg. The floor is frictionless, so the only horizontal force on the lower block is this friction: its largest acceleration is a_max = μ_s mg/(2m) = 0.5μ_s g. Moving together, both blocks have this acceleration: F_max = (3m)(0.5μ_s g) = 1.5μ_s mg. (Force equated with the maximum friction between the blocks: 1.0μ_s mg. Common acceleration capped at μ_s g: (3m)(μ_s g) = 3.0μ_s mg. Normal force between the blocks taken as the weight of both, 3mg: a_max = 3μ_s mg/(2m) = 1.5μ_s g, so F = (3m)(1.5μ_s g) = 4.5μ_s mg.) Checked with sympy for general masses: F_max = μ_s m(M + m)g/M with M = 2m.

CED 2.7.B.2.i · Read this in Fix

1. A0.72 mg²/b
   A student who uses μ_k mg as the threshold for slipping starts the motion at t = 0.40mg/b and picks this: integrating (bt − 0.40mg)/m from 0.40mg/b to 1.6mg/b gives 0.72mg²/b. Before sliding, static friction can rise to 0.80mg, so the block starts later, at 0.80mg/b.
2. B0.96 mg²/b
   A student who drops friction once the block breaks free integrates bt/m from 0.80mg/b to 1.6mg/b and picks this. Kinetic friction 0.40mg keeps opposing the sliding, which lowers the speed by 0.32mg²/b.
3. C0.32 mg²/b
   A student who keeps the friction at μ_s mg = 0.80mg after sliding begins picks this. Once the block slides, the friction is kinetic, 0.40mg, which is smaller, so the block gains more speed than this.
4. D0.64 mg²/b Correct
   The block stays at rest until bt reaches μ_s mg = 0.80mg, at t₁ = 0.80mg/b. After that, kinetic friction 0.40mg opposes the sliding, so a = (bt − 0.40mg)/m. Integrating from t₁ to 1.6mg/b gives v = 0.96mg²/b − 0.32mg²/b = 0.64mg²/b.

Working Static friction holds the block at rest until bt reaches μ_s mg = 0.80mg, at t₁ = 0.80mg/b; the given time is 2t₁. After t₁ the block slides and kinetic friction 0.40mg opposes the motion: a(t) = (bt − 0.40mg)/m. v = ∫ from t₁ to 2t₁ of (bt/m − 0.40g) dt = (b/2m)(3t₁²) − 0.40g t₁ = 0.96mg²/b − 0.32mg²/b = 0.64mg²/b. In general, v = μ_s mg²(3μ_s − 2μ_k)/(2b). (Sliding taken to start when bt = μ_k mg = 0.40mg: v = mg²(2μ_s − μ_k)²/(2b) = 0.72mg²/b. No friction once sliding: 3μ_s²mg²/(2b) = 0.96mg²/b. Friction kept at μ_s mg after sliding starts: μ_s²mg²/(2b) = 0.32mg²/b.) Checked with sympy.

CED 2.7.B.2.ii · Read this in Fix