The oxygen cylinder ramp
A hospital is moving a full oxygen cylinder from the ground to a loading bay whose floor is 0.90 m above ground. Two wooden planks can serve as a ramp: plank A is 3.6 m long and plank B is 2.4 m long. A porter can push along the plank with a maximum steady force of 85 N. The cylinder and trolley together have a mass of 24 kg; rolling friction along the plank is a constant 15 N opposing motion; take g = 10 m/s². The porter wants to do the job alone in one steady push, without the cylinder running back, and the cylinder must stay upright; the bay edge gives only 0.30 m of flat overlap for the plank. A supervisor says plank B is better because 'a shorter ramp means less work'. Just before the move, the stores assistant reports that the bay floor may actually be 1.20 m high, not 0.90 m.
1. Using the stated forces, decide whether the porter can move the cylinder up plank A and plank B alone. Show the force balance along the plank and state which plank should be used.
Read the worked answer
Take the plank as the axis. Weight has a component mg sinθ down the slope; friction adds 15 N up the slope against motion, so the push needed is F = mg sinθ + 15.
- Plank A: sinθ = 0.90/3.6 = 0.25. mg sinθ = 24 × 10 × 0.25 = 60 N. F = 60 + 15 = 75 N. Below the 85 N limit, with 10 N to spare.
- Plank B: sinθ = 0.90/2.4 = 0.375. mg sinθ = 90 N. F = 90 + 15 = 105 N. That is 20 N above the limit, so the porter cannot even start the load in a controlled way; if he tries, the cylinder may stall and run back.
Decision: use plank A. The margin is only about 12%, so the plank must be clean and the 0.30 m overlap is a minimum - a plank that slips off the edge turns a controlled push into a fall. Plank A is also the gentler angle (about 14.5° against 22°), which reduces the risk of an upright cylinder tipping.
2. The supervisor says plank B means less work. Evaluate this claim using work and energy, and comment on efficiency.
Read the worked answer
Gravitational work depends only on the height gained: mgh = 24 × 10 × 0.90 = 216 J, whatever the ramp length. Work done by the porter is F × L: A = 75 × 3.6 = 270 J; B = 105 × 2.4 = 252 J. So the supervisor is right that the shorter ramp needs slightly less effort-work, but wrong in the way that matters: on B the force is 105 N, beyond the porter's limit. Ramp length trades force against distance - that is the mechanical advantage of an inclined plane.
Efficiency: A = 216/270 = 80%; B = 216/252 = 86%. The longer ramp is less efficient because friction acts over 3.6 m instead of 2.4 m (friction work 54 J against 36 J). 'Longer ramp, easier' is true for force, not for energy; the practical choice is set by the force limit and safety, not by efficiency alone.
3. If the bay floor is really 1.20 m high, does the decision change? Give the numbers and one safe alternative.
Read the worked answer
Recalculate with h = 1.20 m. Plank A: sinθ = 1.20/3.6 = 0.333; mg sinθ = 240 × 0.333 = 80 N; F = 80 + 15 = 95 N, which exceeds 85 N. The porter now falls 10 N short, so the single-handed plan fails. Plank B: sinθ = 0.5; F = 120 + 15 = 135 N, clearly impossible.
A longer plank would work: for F ≤ 85 N we need 240 × 1.20/L + 15 ≤ 85, i.e. 288/L ≤ 70, so L ≥ 4.1 m. A 4.5 m plank gives sinθ = 0.267 and F = 64 + 15 = 79 N. Equally safe is a second person, a hand winch or a scissor-lift trolley. What must not happen is repeating plank A at 1.20 m 'because it worked at 0.90 m': the 10 N shortfall is exactly the situation where a load stalls mid-ramp and rolls back.