Question 5 (paper Question V) — Crate Sliding Down a 3-4-5 Incline onto a Moving Cart (Part B · Dynamics, equal value): 04-BS-3, December 2019, Q5

Question 5 (paper Question V) — Crate Sliding Down a 3-4-5 Incline onto a Moving Cart (Part B · Dynamics, equal value)

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.

Given. Crate 10 lb$_f$, $v_A=2.5$ ft/s parallel to a 3-4-5 incline (rise:run:length = 3:4:5). Vertical rise A to B (origin) is 15 ft, so incline length $L_{AB}=15/(3/5)=25$ ft and horizontal run $=25(4/5)=20$ ft. $\mu_k=0.2$ on AB. Origin B is 30 ft above the ground; the cart's bed (point C) is 5 ft above the ground, i.e. 25 ft below B.

Find. The horizontal landing distance $x$ (measured from B) where the crate lands on the cart.

Figure 5. Crate slides A to B on a 3-4-5 incline with friction, then falls as a projectile onto the cart at C.

Approach. Use the work-energy principle along the incline (gravity component minus kinetic friction) to find the speed and direction at B, then treat the motion from B to C as free-fall projectile motion under gravity alone.

Speed at B (work-energy on the incline). Normal force $N=W\cos\alpha$ with $\cos\alpha=4/5$; friction $=\mu_kN$ opposes motion. Net accelerating component along the incline: $g(\sin\alpha-\mu_k\cos\alpha)=g(0.6-0.2\times0.8)=0.44g$. $$v_B^2=v_A^2+2(0.44g)L_{AB}=2.5^2+2(0.44)(32.2)(25)=714.6\ \Rightarrow\ v_B=\boxed{26.73\text{ ft/s}}$$
Velocity components at B. Still parallel to the 3-4-5 incline: $$v_{Bx}=v_B(4/5)=21.39\text{ ft/s},\qquad v_{By}=-v_B(3/5)=-16.04\text{ ft/s}$$
Projectile motion, B (origin) to landing on the cart. With $y(t)=v_{By}t-\tfrac12 gt^2$ and the cart bed at $y=-25$ ft: $$-25=-16.04t-16.1t^2\ \Rightarrow\ 16.1t^2+16.04t-25=0\ \Rightarrow\ \boxed{t=0.844\text{ s}}$$
Horizontal landing distance. $$x=v_{Bx}\,t=21.39(0.844)=\boxed{18.0\text{ ft}}$$

Final Results — Question 5
Quantity	Value
$L_{AB}$	25 ft
$v_B$	26.73 ft/s
Time of flight $t$	0.844 s
Landing distance $x$	18.0 ft

Topic: Work-energy principle on an incline with friction, followed by projectile motion.

Key relations: $\tfrac12mv_A^2+U_{1\to2}=\tfrac12mv_B^2$ with $U_{1\to2}=W(\sin\alpha-\mu_k\cos\alpha)L$; projectile kinematics $x=v_xt$, $y=v_yt-\tfrac12gt^2$.

Why this works. The incline constrains the crate to move in a straight line, so a scalar work-energy equation along that line is faster than resolving Newton's second law into normal/tangential components and integrating; once the crate leaves the incline at B it is a free particle under gravity alone, and the two motion phases only share the single state (speed and direction) at B.

Common pitfall: using $g=9.81$ (SI) instead of $g=32.2$ ft/s² because the problem is posed in US customary units; also forgetting that the friction force depends on $\cos\alpha$ (the true incline normal), not on the crate's weight directly.

Source: Hibbeler, Engineering Mechanics: Dynamics, Ch. 14 (Kinetics of a Particle: Work and Energy), Ch. 12 (projectile motion).

Question 5 of 6: Question 5 (paper Question V) — Crate Sliding Down a 3-4-5 Incline onto a Moving Cart (Part B · Dynamics, equal value)

Question 5 (paper Question V) — Crate Sliding Down a 3-4-5 Incline onto a Moving Cart (Part B · Dynamics, equal value)