A hoist is a block and tackle: a rope threaded back and forth between a fixed pulley block and a movable one, so that the load hangs from several strands at once instead of from a single rope. Cranes on building sites, theatre fly systems, sailing-boat mainsheets and garage engine lifts are all the same machine, and all of them exist for the same reason — a person can pull hard for a short time but cannot pull with a force of several hundred newtons at all.
The principle is that the load’s weight is shared equally between the strands that support the movable block. With five such strands, each carries a fifth of the weight, and since the free end of the rope is simply the last of those strands, the hand pulls only a fifth as hard. Nothing is gained for nothing: to raise the load by 20 cm, each of the five strands must shorten by 20 cm, so a full metre of rope has to be pulled through. The machine multiplies force and divides distance, and the product of the two — the work — is what it cannot change. What a real hoist can do is lose some of that work to friction in the axles and to the weight of the movable block it must lift along with the load, and the fraction that survives is its efficiency.
In this laboratory you will hang a 1 N weight from a five-strand hoist, pull on a dynamometer until the load rises at constant speed, and read the force required. You will repeat the measurement eight more times, adding a newton each time up to 9 N, and for every trial compute the ratio of the load to the pulling force. Newton’s second law applied to the load at constant velocity then gives the relation between the weight and the rope tension, and comparing the work you put in with the potential energy the load gained gives the efficiency of the machine.
Educational Goals
Understanding how a block and tackle multiplies force
- Count the strands supporting the movable block and predict, before measuring anything, that the pulling force will be the load divided by that number.
- Explain why the ideal mechanical advantage is a count of strands and not a property of the pulleys themselves.
Applying Newton’s laws to a system in equilibrium
- Draw the free-body diagram of the movable block and write ΣF = 0 for a load rising at constant speed, obtaining Fg = 5T with T equal to the applied force.
- Recognise that “constant speed” is the condition that makes the analysis a statics problem, and that a jerk at the start is a different, harder problem.
Measuring force and displacement
- Use a dynamometer to read a steady pulling force while keeping the motion uniform, and judge the reading against the instrument’s stated precision rather than its printed digits.
- Measure the length of rope drawn in against the height gained by the load, and use their ratio as an independent check on the strand count.
Work, energy and efficiency
- Compute the work put in as W = FΔx and the potential energy gained as Ep = FgΔy, and identify the difference as work lost within the machine.
- Express the efficiency two equivalent ways — as Ep/W and as the ratio of actual to ideal mechanical advantage — and show that they are the same quantity.
Interpreting a trend across a series of loads
- Plot the measured force against the load for all nine trials and interpret both the slope and the intercept, the second of which is the overhead the machine carries whatever the load.
- Decide from the data whether the losses scale with the load or are fixed, and explain what each case implies for the efficiency of a heavily loaded crane.
Connecting the machine to its uses
- Relate the strand count to real equipment — a theatre counterweight line, a crane block, a vehicle recovery tackle — and explain the trade-off between the force saved and the rope that must be hauled.
Protocol
- Suspend a weight of 1 N from the assembly of movable pulleys (black ring).
- Bring one hand to the end of the dynamometer in order to pull on the hook; which will have the effect of lifting the load at a constant speed.
- The force indicated by the dynamometer is recorded in the results table.
- Repeat eight other times steps 1 and 2 while increasing each time the suspended weight by 1 N.
- For each test; calculate the ratio of the weight of the load Fg to the required driving force F.
- Apply Newton’s second law to the load in order to obtain a relationship between the weight and the tension in the rope.
- Calculate the mechanical work that was done to lift the load to a height of 20 cm at constant speed.
- Determine the amount of potential energy acquired by the load.
- Determine the energy efficiency of the hoist.
Anticipated Outcomes
Nine loads, from 1 N to 9 N in one-newton steps, are lifted in turn and the force needed to raise each at constant speed is read from the dynamometer. The expected pattern is a straight line: the force required rises in proportion to the load, at about a fifth of its value, because five strands share the weight. The laboratory publishes two of the nine trials, and they are reproduced here as the values of record together with everything that can be derived from them.
| Trial | Load Fg | Measured force F | Ratio Fg/F | Rope drawn in | Work in W = FΔx | Energy gained Ep = FgΔy | Efficiency |
|---|---|---|---|---|---|---|---|
| Light-load example | 2.00 N | 0.40 N | 5.00 | 1.00 m | 0.40 J | 0.40 J | 100 % |
| Heaviest load | 9.00 N | 2.02 N | 4.46 | 1.00 m | 2.02 J | 1.80 J | 89 % |
Why the rope travels five times as far. Each of the five strands supporting the movable block must shorten by the height the load rises. Raising the load by Δy = 0.200 m therefore takes in Δx = 5 × 0.200 = 1.00 m of rope, and that displacement ratio is the ideal mechanical advantage stated a second way: IMA = Δx/Δy = 5. This is the more reliable of the two ways to establish the strand count, because it involves no force measurement at all — a class that is unsure how many strands the hoist has should measure the rope drawn in for a known lift and divide.
The equilibrium condition. With the load rising at constant speed there is no acceleration, so for the movable block and its load ΣF = 0: the five upward strand tensions balance the weight together with the block’s own weight, 5T = Fg + wblock. The rope is continuous over frictionless pulleys, so the tension is the same everywhere along it and equals the force at the hand, T = F. In the ideal case, with a weightless block and no friction, this reduces to F = Fg/5 — substituting the heaviest load, F = 9.00/5 = 1.80 N, against the 2.02 N actually needed.
| Load Fg | Ideal force Fg/5 | Rope drawn in for a 0.200 m lift | Energy gained Ep | Ideal work in |
|---|---|---|---|---|
| 1.00 N | 0.20 N | 1.00 m | 0.20 J | 0.20 J |
| 2.00 N | 0.40 N | 1.00 m | 0.40 J | 0.40 J |
| 3.00 N | 0.60 N | 1.00 m | 0.60 J | 0.60 J |
| 4.00 N | 0.80 N | 1.00 m | 0.80 J | 0.80 J |
| 5.00 N | 1.00 N | 1.00 m | 1.00 J | 1.00 J |
| 6.00 N | 1.20 N | 1.00 m | 1.20 J | 1.20 J |
| 7.00 N | 1.40 N | 1.00 m | 1.40 J | 1.40 J |
| 8.00 N | 1.60 N | 1.00 m | 1.60 J | 1.60 J |
| 9.00 N | 1.80 N | 1.00 m | 1.80 J | 1.80 J |
The numerical coincidence in that table is worth pointing out to students rather than letting it confuse them: the ideal force in newtons and the potential energy in joules take the same numbers, because the rope is drawn in by exactly 1.00 m and multiplying by one changes nothing. Change the lift height and the coincidence disappears while the physics does not.
What the two measured trials say about the losses. Subtracting the ideal force from the measured one gives the machine’s overhead. At the heaviest load, 5F − Fg = 5 × 2.02 − 9.00 = 1.10 N; at the light-load example, 5 × 0.40 − 2.00 = 0.00 N. Those two results cannot both describe the same machine. The weight of the movable block is a fixed quantity that does not know how heavy the load is, so if it and the axle friction together account for 1.10 N at 9 N, the same 1.10 N must be present at 2 N, which would make the pulling force (2.00 + 1.10)/5 = 0.62 N and the ratio 3.2 rather than 5.0. The light-load figure of exactly 0.40 N gives a ratio of exactly 5.00 and an efficiency of exactly 100 %, which no machine with mass or friction can reach.
Which way should the efficiency go? This is the question the nine-trial series exists to answer, and it has only two defensible answers. If the dominant loss is the weight of the movable block, the overhead is a constant number of newtons, the efficiency is Fg/(Fg + w) and it rises towards 100 % as the load grows — the heavier the load, the less the machine’s own weight matters. If the dominant loss is axle friction, which grows with the tension in the rope, the overhead is a constant fraction and the efficiency is flat across the series. A hoist that becomes less efficient as it is loaded more heavily is not what either mechanism predicts, so a downward trend in the data should be treated as a finding to investigate rather than as the expected result. The plot to make is the measured force against the load: its slope is 1/(5η) for the proportional case and its intercept is w/5 for the fixed-overhead case, and the nine points decide between them.
The energy lost per lift. At the heaviest load the work put in is 2.02 J and the load gains 1.80 J, so 0.22 J stays in the machine, as warmth in the axles and as work done raising the movable block itself. Over a hundred lifts that is 22 J — small here, but the same arithmetic applied to a crane raising a tonne through ten metres puts nearly eleven megajoules into the load and over a megajoule into the block and its bearings, which is why real hoists use ball races and why their efficiency is an economic quantity rather than an academic one.
Summary of Assignment by Grade Range
Grade 9–10
Focus: seeing the force reduction and describing it correctly. Students count the strands supporting the movable block, predict the pulling force for each load before measuring it, and complete the nine-trial table of load against force. They measure the rope drawn in for a 20 cm lift and find that it is a metre, which tells them the machine gives back in distance what it saves in force. The expected conclusions: the hoist reduces the force needed by about five times, it does not reduce the effort in the sense of work, and the measured force is always a little more than a fifth of the load rather than exactly a fifth. Vocabulary to be used correctly: load, tension, mechanical advantage, work, efficiency.
Grade 11
Focus: the quantitative treatment. Students write ΣF = 0 for the movable block, derive Fg = 5T, and compute for every trial the ratio Fg/F, the work in W = FΔx, the energy gained Ep = FgΔy and the efficiency Ep/W. They plot the measured force against the load and extract both the slope and the intercept, then check that the efficiency obtained from the energies agrees with the one obtained from the force ratio divided by five — they are algebraically the same number, and getting two different answers means an arithmetic slip. The target discussion is what the intercept represents physically.
Grade 12 / College Level
Focus: distinguishing the loss mechanisms and bounding the result. Students model the machine two ways — fixed overhead, F = (Fg + w)/5, and proportional loss, F = Fg/(5η) — fit both to the nine points and decide which the data supports. The concluding exercise is to measure the movable block’s weight directly and check whether it accounts for the intercept; if it does not, the remainder is friction and rope stiffness, and its magnitude can be quoted.
Laboratory essentials
Instruments
- Hoist (five-strand block and tackle, with movable pulley block)
- Dynamometer
- Weights, 1 N to 9 N in 1 N steps
- 50 cm ruler
Products
None — this laboratory uses no chemical reagents.