Question 4 (paper Question IV) — Motor Power with Friction (Part B · Dynamics, 20 marks): 04-BS-3, December 2013, Q4

Question 4 (paper Question IV) — Motor Power with Friction (Part B · Dynamics, 20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check — the figure caption describes the cable running from the motor, around a pulley mounted on the crate, then around a second pulley on the platform, and back to a tie-off on the crate. Counting the rope legs that pull on the crate (both sides of the crate-mounted pulley, plus the direct tie-off) gives a 3:1 purchase (v_cable = 3v_crate, force on crate = 3F(t)) — the standard reading of this pulley arrangement and the configuration explicitly called a "mechanical advantage" in the caption.

Given. Crate mass m = 175 kg; μ_s = 0.3, μ_k = 0.2; cable (motor) force F(t) = 8t² + 20 N; 3:1 pulley purchase between cable and crate.

Given data
Quantity	Value
m	175 kg
μ_s, μ_k	0.3, 0.2
F(t)	8t² + 20 N
Purchase	3:1

Find. The power output of the motor at t = 5 s.

Figure 4 — motor, platform pulley, crate pulley and tie-off giving a 3:1 cable purchase on the crate.

Approach. Find when the crate first breaks static friction, then integrate Newton's second law (with the time-varying driving force and constant kinetic friction) from that instant to t = 5 s to get the crate's speed, then convert to cable speed and multiply by the cable force for power.

Time the crate starts to move — driving force 3F(t) first equals the maximum static friction $\mu_s mg$: $$\mu_s m g = 0.3(175)(9.81) = 515.03\ \text{N} = 3(8t_0^2+20) \Rightarrow t_0 = \boxed{4.354\ \text{s}}$$
Equation of motion for $t \gt t_0$ (kinetic friction now, constant): $$m\dfrac{dv}{dt} = 3(8t^2+20) - \mu_k m g = 24t^2+60-343.35$$
Integrate from t₀ to 5 s to get the crate's speed: $$v_{crate}(5) = \int_{4.354}^{5}\dfrac{24t^2-283.35}{175}\,dt = \boxed{0.895\ \text{m/s}}$$
Cable force and speed at t = 5 s, then power: $$F(5) = 8(5)^2+20 = \boxed{220\ \text{N}}, \qquad v_{cable}(5) = 3\,v_{crate}(5) = \boxed{2.68\ \text{m/s}}$$ $$P = F(5)\,v_{cable}(5) = 220(2.68) = \boxed{590.6\ \text{W}}$$

Motor power at t = 5 s
Quantity	Value
t₀ (motion begins)	4.354 s
v_crate(5 s)	0.895 m/s
F(5 s)	220 N
v_cable(5 s)	2.68 m/s
P(5 s)	590.6 W

Topic: Kinetics of a particle with a time-varying force, and power/purchase in pulley systems.

Key relations: $P = \mathbf{F}\cdot\mathbf{v}$ at the point of force application (the motor's cable end); Newton's second law $m\,dv/dt = \sum F$; pulley purchase relates cable speed to load speed.

Why this works: the crate is stationary (governed by static friction) until the driving force exceeds $\mu_s mg$; afterward the motion is governed by kinetic friction and the now time-varying net force, integrated directly since acceleration is not constant.

Common pitfall: computing power as F(5)·v_crate(5) instead of F(5)·v_cable(5) — the motor does work at the speed the CABLE moves, not the speed the crate moves, when a purchase system is present.

Source: Hibbeler, Engineering Mechanics: Dynamics, 14th ed., Ch. 13 (Kinetics of a Particle: Force and Acceleration) & Ch. 14 (Work and Energy).

Question 4 of 6: Question 4 (paper Question IV) — Motor Power with Friction (Part B · Dynamics, 20 marks)

Question 4 (paper Question IV) — Motor Power with Friction (Part B · Dynamics, 20 marks)