CELE Engineering Mechanics — FrictionStudy Notes

Problem

A 200 N block sits on a horizontal surface with μₛ = 0.35. A horizontal force P = 50 N is applied. Determine (a) the friction force, and (b) whether the block moves.

Solution

**Part (a): Find the friction force.** For the block to be in equilibrium horizontally, the friction force must equal the applied force. Friction required for equilibrium: F_equilibrium = P = 50 N Maximum static friction available: F_max = μₛ × N = 0.35 × 200 = 70 N Since F_equilibrium (50 N) < F_max (70 N), static friction can provide the required force. **Answer (a): F = 50 N** (static friction, not at its maximum) **Part (b): Does the block move?** No. Since the applied force (50 N) does not exceed the maximum static friction (70 N), the block remains stationary. The friction force automatically adjusts to exactly 50 N to maintain equilibrium.

Problem

The same 200 N block with μₛ = 0.35 is now subjected to a horizontal force P = 75 N. Find the friction force and the acceleration.

Solution

**Check against maximum friction:** F_max = μₛ × N = 0.35 × 200 = 70 N Applied force P = 75 N Since P (75 N) > F_max (70 N), static friction is insufficient. The block enters kinetic sliding. Assuming μₖ ≈ 0.30 (typical for this material pair, less than μₛ): **Kinetic friction:** Fₖ = μₖ × N = 0.30 × 200 = 60 N **Net force:** F_net = P - Fₖ = 75 - 60 = 15 N **Acceleration:** a = F_net / m = 15 / (200/9.81) = 15 / 20.39 = **0.735 m/s²** **Answer: Friction force = 60 N (kinetic); acceleration = 0.735 m/s² in the direction of P.**

Problem

Two surfaces have μₛ = 0.45. Find the angle of friction φ.

Solution

Using the definition: $$\phi = \arctan(\mu_s) = \arctan(0.45) = 24.23°$$ The total reaction at impending slip makes an angle of **24.23°** with the surface normal.

Problem

A granular soil has μ = 0.6. (a) Find the angle of repose. (b) Will a block rest on a 25° slope? (c) At what angle will it be on the verge of sliding?

Solution

**Part (a): Angle of repose** $$\theta_{\text{repose}} = \arctan(0.6) = 31.0°$$ **Part (b): Block on 25° slope?** Since 25° < 31.0°, the block is **self-locking** and will remain stationary without external support. **Part (c): Angle for impending slip?** The block will be on the verge of sliding exactly at: $$\theta = 31.0°$$ Above this angle, the block slides; below it, the block self-locks.

Problem

A 500 N block rests on a 30° incline with μₛ = 0.25. Find the force P (parallel to the incline) for impending motion up the slope.

Solution

**Given:** - W = 500 N - θ = 30° - μₛ = 0.25 **Free-body diagram:** Incline-aligned axes with W resolved as: - Parallel (down slope): W sin 30° = 500 × 0.5 = 250 N - Perpendicular (into slope): W cos 30° = 500 × 0.866 = 433 N **Normal force:** N = W cos 30° = 433 N **Friction (opposing upward motion, acts down slope):** F = μₛ N = 0.25 × 433 = 108.25 N **Equilibrium parallel to slope (up positive):** $$P = W\sin\theta + \mu N = 250 + 108.25 = 358.25 \text{ N}$$ Or directly: $$P = W(\sin\theta + \mu\cos\theta) = 500(0.5 + 0.25 × 0.866) = 500(0.7165) = 358.25 \text{ N}$$ **Answer: P = 358 N** (parallel to the incline, directed up the slope)

Problem

Same block (500 N, 30° incline, μₛ = 0.25). Find the minimum force P to prevent it from sliding down.

Solution

**Check for self-locking:** $$\tan\theta = \tan 30° = 0.577$$ $$\mu = 0.25$$ Since tan 30° (0.577) > μ (0.25), the block would slide on its own without external support. **Impending motion down the slope:** Friction acts up the slope (opposing downward motion). **Normal force:** N = W cos 30° = 433 N (same as before) **Friction (up the slope):** F = μₛ N = 0.25 × 433 = 108.25 N **Equilibrium parallel to slope (up positive):** $$P + F = W\sin\theta$$ $$P = W\sin\theta - F = 250 - 108.25 = 141.75 \text{ N}$$ Or directly: $$P = W(\sin\theta - \mu\cos\theta) = 500(0.5 - 0.25 × 0.866) = 500(0.2835) = 141.75 \text{ N}$$ **Answer: P = 142 N** (parallel to the incline, directed up the slope to prevent sliding) **Note:** Compare to Example 3.1: pushing up requires 358 N, but holding against downslope motion requires only 142 N. The friction helps in the holding case and opposes in the pushing case.

Problem

A 400 N block sits on a 20° incline with μₛ = 0.3. A force P = 150 N is applied at 15° above the incline surface (measured from the incline). Determine whether the block is in equilibrium, on the verge of sliding up, or on the verge of sliding down.

Solution

**Resolve the applied force:** - Parallel (up slope): P∥ = 150 cos 15° = 150 × 0.9659 = 144.9 N - Perpendicular (away from slope): P⊥ = 150 sin 15° = 150 × 0.2588 = 38.8 N **Normal force (perpendicular to slope, into surface is positive):** The weight component perpendicular: W cos 20° = 400 × 0.9397 = 375.9 N (into slope) The applied force component perpendicular: 38.8 N (away from slope) $$N = W\cos\theta - P\sin 15° = 375.9 - 38.8 = 337.1 \text{ N}$$ **Maximum static friction:** $$F_{\max} = \mu N = 0.3 × 337.1 = 101.1 \text{ N}$$ **Weight component parallel (down slope):** $$W\sin 20° = 400 × 0.342 = 136.8 \text{ N}$$ **Net parallel force (without friction):** $$P∥ - W\sin\theta = 144.9 - 136.8 = 8.1 \text{ N (up the slope)}$$ **Since P∥ > W sin θ, the block tends to slide UP.** **Friction opposes upward motion (acts down slope):** For equilibrium: $$P∥ = W\sin\theta + F$$ $$144.9 = 136.8 + F$$ $$F = 8.1 \text{ N}$$ **Since required friction (8.1 N) < maximum static friction (101.1 N), the block is in equilibrium** at rest—it will not slide. **Check: If we slowly reduced P, at what point would the block start to slide down?** This would occur when P∥ = W sin θ − μN, requiring additional analysis. But with P = 150 N as given, the block is stable.

Problem

A 10 kN block rests against a smooth (frictionless) vertical wall. A wedge with angle α = 10° and μ = 0.2 is driven horizontally beneath the block to lift it. Assume the block is about to be lifted (impending motion of wedge into position). Find (a) the normal force exerted by the wedge on the block, and (b) the driving force P.

Solution

**Setup:** The block has weight W = 10 kN acting downward. It contacts the wall (frictionless, horizontal reaction H) and the inclined wedge surface (normal force N₁ and friction F₁). **Free-body diagram of the block (at impending motion):** - Weight: W = 10 kN (downward) - Wall reaction: H (horizontal, away from wall) - Wedge normal: N₁ (perpendicular to wedge, at angle 10° from vertical) - Friction on wedge: F₁ = μN₁ = 0.2N₁ (along wedge surface, up the incline since wedge moves horizontally forward) **Decompose N₁ and F₁:** The normal force N₁ makes angle (90° − α) = 80° with the horizontal, or equivalently, α = 10° from the vertical. - Horizontal component of N₁: N₁ sin α = N₁ sin 10° = 0.1736N₁ - Vertical component of N₁: N₁ cos α = N₁ cos 10° = 0.9848N₁ The friction force F₁ is along the wedge surface (parallel to incline), at angle α = 10° above horizontal: - Horizontal component of F₁: F₁ cos α = 0.2N₁ cos 10° = 0.1970N₁ (forward/to the right) - Vertical component of F₁: F₁ sin α = 0.2N₁ sin 10° = 0.0347N₁ (upward) **Equilibrium of block:** *Vertical:* $$N₁\cos\alpha + F₁\sin\alpha = W$$ $$0.9848N₁ + 0.0347N₁ = 10$$ $$1.0195N₁ = 10$$ $$N₁ = 9.81 \text{ kN}$$ *Horizontal:* $$H = N₁\sin\alpha + F₁\cos\alpha = 0.1736 × 9.81 + 0.1970 × 9.81 = 1.703 + 1.933 = 3.636 \text{ kN}$$ **Free-body diagram of the wedge:** - Applied force P (horizontal, to the right) - Reaction from block: N₁ = 9.81 kN (perpendicular to wedge surface, pointing into the wedge) and F₁ = 0.2 × 9.81 = 1.962 kN (along wedge surface, pointing down the incline relative to wedge motion) - Normal from ground: N₂ (vertical, upward) - Friction from ground: f₂ = μN₂ = 0.2N₂ (horizontal, opposing wedge motion) - Weight of wedge (often neglected if light): assume negligible **Decompose the reaction forces on the wedge from the block:** Reaction from block acts opposite to action on block. - N₁ points perpendicular into the wedge (at angle α below horizontal): - Horizontal: 9.81 sin 10° = 1.703 kN (to the right, helps P) - Vertical: 9.81 cos 10° = 9.661 kN (downward) - F₁ points up the incline relative to the wedge surface (friction opposes wedge sliding), which is down and to the left at angle α below the incline normal: - Horizontal: 1.962 cos 10° = 1.933 kN (to the left, opposes P) - Vertical: 1.962 sin 10° = 0.341 kN (downward) **Equilibrium of wedge:** *Vertical:* $$N₂ = 9.661 + 0.341 = 10.002 \text{ kN} ≈ 10 \text{ kN}$$ (as expected, wedge supports the block's weight) *Horizontal:* $$P = N₁\sin\alpha + F₁\cos\alpha + f₂$$ $$P = 1.703 + 1.933 + 0.2 × 10 = 1.703 + 1.933 + 2.0 = 5.636 \text{ kN}$$ **Answers:** - **(a) Normal force from wedge on block: N₁ = 9.81 kN** (perpendicular to wedge surface, at 10° from vertical) - **(b) Driving force: P = 5.64 kN** **Check for self-locking:** φ = arctan(0.2) = 11.31°. Since α = 10° < 2φ = 22.62°, the wedge is self-locking; it will not slide back out if P is removed.

Problem

A wedge with angle α = 8° has μ = 0.15 on both contact surfaces. Will this wedge self-lock?

Solution

**Self-locking criterion for a wedge with friction on both surfaces:** $$\alpha < 2\phi$$ where φ = arctan(μ). **Calculate:** $$\phi = \arctan(0.15) = 8.53°$$ $$2\phi = 2 × 8.53° = 17.06°$$ **Compare:** α = 8° < 2φ = 17.06° **Answer: Yes, the wedge is self-locking.** Once driven in place, the friction on both the inclined surface (wedge–block) and the lower surface (wedge–ground) is sufficient to prevent the wedge from being pushed back out by the block's weight or reaction forces. This is desirable for stabilizing structures or maintaining position without continuous applied force.

Problem

A rope wraps 180° (half turn) around a fixed drum with μ = 0.25. The tight side carries a tension of 1200 N. Find the slack-side tension at impending slip.

Solution

**Given:** - Contact angle: β = 180° = π rad ≈ 3.14159 rad - μ = 0.25 - T_tight = 1200 N **Apply the Capstan equation:** $$\frac{T_{\text{tight}}}{T_{\text{slack}}} = e^{\mu\beta}$$ $$\frac{1200}{T_{\text{slack}}} = e^{0.25 × 3.14159} = e^{0.7854} = 2.193$$ $$T_{\text{slack}} = \frac{1200}{2.193} = 547 \text{ N}$$ **Answer: T_slack = 547 N** **Physical interpretation:** With only a 180° wrap, the slack tension is about 45% of the tight tension. This is because the exponential factor is not very large; longer wraps produce more dramatic ratios.

Problem

A ship's hawser (rope) wraps 1.5 times around a capstan (fixed post) on the dock. The rope has μ ≈ 0.5 with the wood capstan. A dock worker pulls on the free end with force P = 200 N. What maximum tension can be held by a rope attached to the ship?

Solution

**Given:** - Number of wraps: 1.5 turns = 1.5 × 2π = 3π rad - Contact angle: β = 3π ≈ 9.4248 rad - μ = 0.5 - Applied force (slack side): T_slack = 200 N - Find: T_tight (tension holding the ship) **Apply the Capstan equation:** $$\frac{T_{\text{tight}}}{T_{\text{slack}}} = e^{\mu\beta}$$ $$T_{\text{tight}} = T_{\text{slack}} × e^{0.5 × 9.4248} = 200 × e^{4.7124} = 200 × 111.3 = 22,260 \text{ N} ≈ 22.3 \text{ kN}$$ **Answer: T_tight ≈ 22.3 kN** (about 22 times the applied force!) **Practical note:** This is why ships are secured to dock capstans with wraps. A single worker (200 N pull) can hold a ship exerting thousands of newtons of tension, thanks to the exponential capstan effect. Each additional wrap dramatically increases the mechanical advantage.

Problem

A flat belt drives a pulley system. The driving pulley (motor side) has tension T₁ = 3000 N, and the driven pulley experiences resistance causing tension T₂ = 1500 N. The belt wraps 120° around the motor pulley. What minimum coefficient of friction is needed to prevent slip?

Solution

**Given:** - Contact angle: β = 120° = 120 × π/180 = 2.0944 rad - T_tight = T₁ = 3000 N (motor side, higher tension) - T_slack = T₂ = 1500 N (driven side, lower tension) - Find: minimum μ to prevent slip **Rearrange the Capstan equation to solve for μ:** $$\frac{T_{\text{tight}}}{T_{\text{slack}}} = e^{\mu\beta}$$ $$\ln\left(\frac{T_{\text{tight}}}{T_{\text{slack}}}\right) = \mu\beta$$ $$\mu = \frac{1}{\beta} \ln\left(\frac{T_{\text{tight}}}{T_{\text{slack}}}\right) = \frac{1}{2.0944} \ln\left(\frac{3000}{1500}\right)$$ $$\mu = \frac{1}{2.0944} \ln(2) = \frac{0.6931}{2.0944} = 0.331$$ **Answer: μ_min ≈ 0.33** **Design implication:** To safely drive this pulley system, the belt–pulley material pairing must have a coefficient of friction of at least 0.33. In practice, engineers add a safety factor (e.g., require μ ≥ 0.5) to account for belt wear, contamination, and dynamic effects.

Problem

A uniform 400 N ladder leans against a smooth wall. The floor has μₛ = 0.35. Find the minimum angle with the horizontal at which the ladder will not slip.

Solution

**Apply the ladder formula:** $$\theta_{\min} = \arctan\left(\frac{1}{2\mu}\right) = \arctan\left(\frac{1}{2 × 0.35}\right) = \arctan\left(\frac{1}{0.70}\right)$$ $$\theta_{\min} = \arctan(1.4286) = 55.01°$$ **Answer: θ_min ≈ 55°** **Interpretation:** If the ladder is placed at an angle of 55° or steeper (measured from the horizontal), it will not slip. Below 55°, friction at the floor is insufficient to support the ladder's weight, and the base will slide away from the wall. **Practical check:** A 55° angle is fairly steep. Most people intuitively place ladders at 50°–60°, which aligns with safety recommendations.

Problem

The same 400 N ladder (from Example 6.1) now has a 800 N person standing 0.75 m from the base. The ladder is 5 m long. Assuming the wall remains smooth and the floor has μₛ = 0.35, find the new minimum angle to prevent slip.

Solution

**Given:** - Ladder weight: W = 400 N (acts at midpoint, 2.5 m from base) - Person weight: P = 800 N (acts at 0.75 m from base) - Ladder length: L = 5 m - μₛ = 0.35 - Find: θ_min **Vertical equilibrium:** $$N_f = W + P = 400 + 800 = 1200 \text{ N}$$ **Moment about the base (clockwise positive):** Wall normal force (horizontal, at height L sin θ) creates a clockwise moment: $$M_{N_w} = N_w × L\sin\theta$$ Ladder weight (at horizontal distance 2.5 cos θ from base): $$M_W = 400 × 2.5\cos\theta = 1000\cos\theta$$ Person weight (at horizontal distance 0.75 cos θ from base): $$M_P = 800 × 0.75\cos\theta = 600\cos\theta$$ Equilibrium: $$N_w × L\sin\theta = 1000\cos\theta + 600\cos\theta$$ $$N_w × 5\sin\theta = 1600\cos\theta$$ $$N_w = \frac{1600\cos\theta}{5\sin\theta} = \frac{320\cos\theta}{\sin\theta} = 320\cot\theta$$ **At impending slip (F = μ N_f):** $$N_w = F = \mu N_f = 0.35 × 1200 = 420 \text{ N}$$ **Solve for θ:** $$320\cot\theta = 420$$ $$\cot\theta = \frac{420}{320} = 1.3125$$ $$\tan\theta = \frac{1}{1.3125} = 0.7619$$ $$\theta_{\min} = \arctan(0.7619) = 37.3°$$ **Answer: θ_min ≈ 37.3°** **Comparison:** With the person on the ladder (37.3°), the minimum angle is significantly steeper than without (55°). The added weight increases friction demand, and the person's position (away from the wall) shifts the moment balance. This is why ladders must be placed at steep angles when carrying loads.

Problem

Verify the result from Example 6.1 by computing the friction force and normal force at the base for the ladder at θ = 55°.

Solution

**Given:** - Ladder weight: W = 400 N - θ = 55° - μₛ = 0.35 **Vertical equilibrium:** $$N_f = W = 400 \text{ N}$$ **Moment about base:** $$N_w × L\sin\theta = W × \frac{L}{2}\cos\theta$$ $$N_w × L\sin 55° = 400 × \frac{L}{2}\cos 55°$$ $$N_w × 0.8192 = 400 × 0.5 × 0.5736$$ $$N_w × 0.8192 = 114.7$$ $$N_w = 140.1 \text{ N}$$ **Horizontal equilibrium:** $$F = N_w = 140.1 \text{ N}$$ **Check against maximum static friction:** $$F_{\max} = \mu N_f = 0.35 × 400 = 140 \text{ N}$$ **Verification:** F ≈ F_max (140.1 ≈ 140 N), confirming that at θ = 55°, the ladder is on the verge of slipping. This matches the theoretical prediction θ_min = 55°.

Problem

A 50 kN block sits on a smooth 25° incline. To hold it in place, a wedge with angle 8° and μ = 0.2 is placed under the block (in contact with the incline surface). The wedge rests on rough ground with μ = 0.2. Find the horizontal driving force P required to insert the wedge and hold the block.

Solution

This problem combines incline and wedge analysis. Key steps: 1. **Block equilibrium on smooth incline (perpendicular to incline surface):** - Normal from incline on block: N_block = W cos 25° = 50 × 0.906 = 45.3 kN 2. **Wedge geometry:** The wedge is placed with its inclined face against the 25° incline. The wedge angle is 8°, so the wedge's inclined surface makes an angle of 25° + 8° = 33° with the horizontal. 3. **Block equilibrium with wedge (free-body diagram of block):** - Weight: 50 kN down - Incline normal: N_incline (perpendicular to incline, but reduced now that wedge shares load) - Wedge normal: N_1 (perpendicular to wedge surface at 33° from horizontal) - Incline friction: F_incline (along incline, opposes block motion) - Wedge friction: F_1 = μN_1 = 0.2 N_1 (along wedge surface, opposes wedge motion) This becomes complex; for brevity, assume the wedge fully supports the block (N_incline → 0). 4. **Block on wedge (impending motion up the incline):** Similar to the block on incline analysis, but now the incline is replaced by the wedge surface at 33°: $$N_1 = W\cos 33° = 50 × 0.839 = 41.95 \text{ kN}$$ $$F_1 = 0.2 × 41.95 = 8.39 \text{ kN}$$ 5. **Wedge equilibrium (impending motion rightward):** - Reaction from block: N_1 perpendicular to wedge (at 33° from horizontal), F_1 along wedge - Applied force: P (horizontal) - Ground normal: N_2 = W = 50 kN (approximately; neglecting wedge weight) - Ground friction: F_2 = 0.2 N_2 = 10 kN (horizontal, opposes wedge motion) **Horizontal equilibrium of wedge:** $$P = N_1\sin 33° + F_1\cos 33° + F_2$$ $$P = 41.95 × 0.545 + 8.39 × 0.839 + 10 = 22.86 + 7.04 + 10 = 39.9 \text{ kN}$$ **Answer: P ≈ 40 kN** **Note:** This is a simplified analysis; a full solution would account for all interactions and friction on multiple surfaces. The key insight is that wedge angle, friction on both surfaces, and the geometry combine to determine the required driving force.