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Wheelbarrow as a lever: torque, advantage, and the lift

SPH4C Grade 12 Physics (College) · Mechanical Systems

A landscaper's wheelbarrow is loaded with 68.0 kg of patio stones. The load's centre of gravity sits 45.0 cm behind the wheel axle, and the lifting grips on the handles are 1.35 m behind that same axle. The barrow is lifted just clear of the ground, handles level, so both the load and the lift act vertically.

Given
  • m = 68 kg — Mass of the load
  • d_l = 45 cm — Axle to the load's centre
  • d_e = 1.35 m — Axle to the lifting grips
Determine
  1. (a)the weight of the load
  2. (b)the torque the load exerts about the axle
  3. (c)the mechanical advantage of the barrow
  4. (d)the upward force the landscaper must apply at the grips
Step 1 of 4(a) · solve for Weight

Torque needs a force, and 68.0 kg is not a force — it is a mass. Multiplying by g first is the step students skip when they write τ = 0.45 × 68 and get an answer in units no wrench has ever read.

Rearranged for W
W=mgW = m g
Your values, in your units
W=(68 kg) (9.80665 m/s2)W = \left( 68\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)
Answer
W=666.85 NW = 666.85\ \text{N}

Carried onward at full precision, not this rounded figure.

Open the Weight (W = mg) solver →

Step 2 of 4(b) · solve for Torque

The load pulls straight down while its lever arm runs horizontally to the axle, so θ = 90° and sin θ = 1 — the full 45.0 cm works. The centimetres convert to 0.450 m here; leave them and the torque inflates a hundred-fold.

Rearranged for τ
τ=rFsin⁡θ\tau = r F \sin\theta
666.85 Ncarried from step 1
Your values, in your units
τ=(45 cm) (666.852 N) sin⁡(90 ∘)\tau = \left( 45\ \text{cm} \right) \, \left( 666.852\ \text{N} \right) \, \sin \left( 90\ ^{\circ} \right)
Converted to base units
τ=(0.45 m) (666.852 N) sin⁡(90 ∘)\tau = \left( 0.45\ \text{m} \right) \, \left( 666.852\ \text{N} \right) \, \sin \left( 90\ ^{\circ} \right)
Answer
τ=300.08 N⋅m\tau = 300.08\ \text{N}{\cdot}\text{m}

Carried onward at full precision, not this rounded figure.

Open the Torque solver →

Step 3 of 4(c) · solve for Mechanical advantage

Effort arm over load arm, and the wheelbarrow is a class-2 lever: the fulcrum is the WHEEL, not your hands, so the load sits between axle and grips and the ratio 1.35/0.450 = 3 comes out greater than one. Invert it and the machine would be making your job harder.

Rearranged for MA
MA=dedlMA = \frac{d_e}{d_l}
Your values, in your units
MA=(1.35 m)(45 cm)MA = \frac{\left( 1.35\ \text{m} \right)}{\left( 45\ \text{cm} \right)}
Converted to base units
MA=(1.35 m)(0.45 m)MA = \frac{\left( 1.35\ \text{m} \right)}{\left( 0.45\ \text{m} \right)}
Answer
MA=3MA = 3

Carried onward at full precision, not this rounded figure.

Open the Mechanical Advantage of a Lever solver →

Step 4 of 4(d) · solve for Applied force

Balance the see-saw: the lift at 1.35 m must supply the same torque the load imposes at 0.450 m, so F = τ/r. The answer must equal the 667 N weight divided by step 3's advantage of 3 — if the two routes disagree, one of them is wrong.

Rearranged for F
F=τrsin⁡θF = \frac{\tau}{r \sin\theta}
300.08 N·mcarried from step 2
Your values, in your units
F=(300.083 N⋅m)(1.35 m) sin⁡(90 ∘)F = \frac{\left( 300.083\ \text{N}{\cdot}\text{m} \right)}{\left( 1.35\ \text{m} \right) \, \sin \left( 90\ ^{\circ} \right)}
Answer
F=222.28 NF = 222.28\ \text{N}

Carried onward at full precision, not this rounded figure.

Open the Torque solver →

Therefore the 68.0 kg load weighs 667 N and twists the barrow about its axle with 300 N·m, but the 3-to-1 lever built into the frame means the landscaper lifts only 222 N — about the weight of 23 kg.

Why this order

A wheelbarrow is the cheapest torque lesson ever built, and the chain follows the logic of the frame. The weight must come first because torque is force times arm and the mass has to become newtons before it can twist anything. The torque in part (b) is what the machine must overcome; the mechanical advantage in part (c) is pure geometry, fixed the day the barrow was welded; and part (d) closes the loop two independent ways — as a torque balance (τ/r) and as weight over advantage (W/MA) — which land on the same 222 N because they are the same statement rearranged. The recurring classroom error is misplacing the fulcrum. Hands feel like the pivot, so students compute MA with arms measured from the grips and get the advantage upside down. The test is to ask what stays still during the lift: the wheel does, the hands do not.

There is no free lunch in the 3-to-1, and part (d)'s 222 N shows what is actually being traded. Lift the grips 30 cm and the load's centre, three times closer to the axle, rises only 10 cm: one third of the force through three times the distance, so the work — force times distance — is untouched. Every lever, gear train and hydraulic ram makes the same bargain, which is why "mechanical advantage" is a force multiplier and never an energy multiplier. Archimedes made exactly this claim about levers in the third century BC; the barrow just packages it with a wheel so the fulcrum can roll to the next job. Doubled up, the reasoning also says where to load a barrow: pile the stones over the axle and d_l shrinks, the load torque shrinks with it, and the same arms lift a heavier load — advice printed in no manual and known to every landscaper.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.