Basic mechanics: power, torque and pressure

November 3, 2026  ·  Indumor

Power is work per unit of time. Torque is force times lever arm. Pressure is force per unit of area. Those three definitions and a handful of constants settle almost everything a shop has to calculate away from the machine tool: the motor for a conveyor, the push of a cylinder, the gearbox ratio and the pull on each leg of a sling.

Force, torque and power

A kilogram hanging from a rope pulls with 9.81 N, because force is mass times gravity. That force at the end of a 0.5 m lever makes a torque of 4.9 N·m. If the lever turns, torque times angular speed is power; since shop speeds are counted in revolutions per minute, the constant 9550 appears: kW = N·m × rpm / 9550. In US units the same relation reads HP = lbf·ft × rpm / 5252.

F = m · g   ·   M = F · r   ·   P = M · ω = M · n9550   ·   HP = T · n5252

F: force (N); m: mass (kg); g = 9.81 m/s²; M: torque (N·m); r: lever arm (m); ω: angular speed (rad/s) = 2·π·n/60; P: power (kW); n: speed (rpm); T: torque in lbf·ft. 9550 is 60000/2π rounded (9549.3).

PMnkWN·mrpm9550constant: 60000 / 2π = 9549.3M = 9550 · P / nn = 9550 · P / MP = M · n / 9550in US units: HP = T · n / 5252, with T in lbf·ft

The table shows what follows: at equal power, torque is inversely proportional to speed. One kilowatt at 3000 rpm delivers 3.2 N·m; the same kilowatt at 100 rpm, downstream of a gearbox, is 95 N·m. That is what gearboxes are for: a little less power comes out than goes in, because of efficiency, and the torque is multiplied by the ratio.

Torque available per kW and per HP at each speed (M = 9549.3 · P / n; T = 5252 · HP / n)
Speed
rpm
Torque per kW
N·m
Torque per kW
kgf·m
Torque per HP
N·m
Torque per HP
lbf·ft
10095.499.73871.2152.52
30031.833.24623.7417.51
50019.101.94814.2410.50
75012.731.2989.497.00
10009.550.9747.125.25
14506.590.6724.913.62
15006.370.6494.753.50
29003.290.3362.461.81
30003.180.3252.371.75

1450 and 2900 rpm are the actual speeds of 4- and 2-pole motors on 50 Hz; 1500 and 3000 are synchronous. On 60 Hz the equivalents are 1750 and 3500 rpm.

Torque, power and speed

Pick what to calculate and enter the other two values; the one being calculated is ignored. It works at the gearbox output as well as at the motor.

Drive ratios and efficiencies

The ratio i of a drive is input speed over output speed, and it equals the diameters or tooth counts the other way round: i = n1/n2 = d2/d1 = z2/z1. In series, ratios multiply and so do efficiencies. A two-stage helical gearbox (0.97 × 0.97) loses 6 %; a 40:1 worm drive can turn a third of the power into heat. Gear formulas are in chapter 15 and belts and chains in chapter 16.

i = n1n2 = d2d1 = z2z1   ·   itotal = i1 · i2 · i3   ·   ηtotal = η1 · η2 · η3   ·   M2 = M1 · i · η

n: speed; d: pulley or pitch diameter; z: number of teeth; subscript 1 is the input (motor) and 2 the output. With i > 1 the speed drops and the torque rises.

Typical efficiency of each drive stage (indicative: it depends on lubrication, load and condition)
StageEfficiency
%
Notes
Spur gears (one pair)97 a 99per stage, lubricated
Helical gears (one pair)96 a 98per stage; the usual industrial gearbox
Bevel gears95 a 98per stage
Worm and wheel50 a 90falls with the ratio: about 85 % at 10:1, 60 % at 60:1 and 50 % at 100:1
V-belt94 a 97new and properly tensioned; lower if it slips
Timing belt (synchronous)97 a 98–
Roller chain96 a 98lubricated
Pair of rolling bearings on a shaft99per shaft
Whole driveproduct of the stagestwo helical stages and a V-belt: 0.97 × 0.97 × 0.95 = 0.89

Worked example · the gearbox of a belt conveyor

A conveyor has a 5.5 kW motor at 1450 rpm and a 20:1 gearbox with 0.94 efficiency (two helical stages). Motor torque: 9550 × 5.5 / 1450 = 36.2 N·m. At the output: 72.5 rpm and 36.2 × 20 × 0.94 = 681 N·m. On a 320 mm drum that torque pulls the belt with 681 / 0.16 = 4256 N (434 kgf, 957 lbf) at 1.21 m/s, and 4256 × 1.21 = 5.2 kW, which is what went in minus the losses. If the belt needs 600 kgf of pull, the motor is short even though the gearbox is right.

Electric motors: speed, frames and current

A three-phase induction motor runs a little below its synchronous speed, which depends on line frequency and the number of poles: 3600 rpm for 2 poles at 60 Hz, 1800 for 4, 1200 for 6 and 900 for 8 (3000, 1500, 1000 and 750 on a 50 Hz grid). Full-load slip runs from 1 % on large motors to 5 % or more on small ones, which is why the nameplate says 1750 or 1450 rather than 1800 or 1500.

Synchronous speed (n = 120 · f / p) and typical full-load speed of three-phase induction motors
Poles60 Hz synchronous
rpm
60 Hz full load
rpm, typical
50 Hz synchronous
rpm
50 Hz full load
rpm, typical
236003400 to 358030002800 to 2980
418001680 to 179015001380 to 1490
612001100 to 11901000900 to 990
8900830 to 890750680 to 740

IEC frames are named after the shaft height in millimeters (a 132 frame has its shaft centerline 132 mm above the feet) and fix the foot pattern, shaft diameter and key: a 7.5 kW 4-pole motor from any maker goes in a 132M frame with a 38 mm shaft. NEMA frames (143T, 182T, 256T and so on) follow a different system, with inch shafts. The current in the table comes from the formula with typical efficiencies and power factors at 380 V; at 460 V multiply by 0.83, at 400 V by 0.95. The nameplate current is the one that sizes the overload relay and the cable.

I = P · 1000√3 · U · cos φ · η  (three-phase)   ·   I = P · 1000U · cos φ · η  (single-phase)

I: line current (A); P: shaft power (kW); U: line-to-line voltage (V); cos φ: power factor; η: efficiency (0.7 to 0.95 depending on size).

IEC frames, 4 poles, 50 Hz: power, shaft diameter and approximate full-load current on 380 V three-phase
IEC framePower
kW
Power
HP
Shaft dia.
mm
Typical η
%
Power factor
typical
Current at 380 V
A
630.180.21165.00.700.6
710.370.51472.00.721.1
800.751.01978.00.761.9
90S1.101.52480.00.782.7
90L1.502.02482.00.803.5
100L2.203.02884.00.814.9
100L3.004.02885.00.826.5
112M4.005.42886.00.828.6
132S5.507.43887.00.8311.6
132M7.5010.13888.00.8415.4
160M11.0014.84289.00.8522.1
160L15.0020.14290.00.8529.8
180M18.5024.84890.50.8636.1
180L22.0029.54891.00.8642.7
200L30.0040.25592.00.8756.9
225S37.0049.66092.50.8769.9
225M45.0060.36093.00.8784.5
250M55.0073.86593.50.87102.7
280S75.00100.67594.00.88137.8
280M90.00120.77594.50.88164.4

Current from I = P / (√3 · U · cos φ · η) with typical η and power factor; the nameplate value is the real one. At 400 V multiply by 0.95, at 460 V by 0.83, at 230 V three-phase by 1.65.

Motor current

Power factor and efficiency are on the nameplate. Size the overload relay and the cable from the nameplate current, not from this estimate.

Hydraulic and pneumatic cylinders

The force of a cylinder is pressure times piston area, nothing more. Extending, the pressure works on the full circle; retracting, on the annulus left once the rod is subtracted, so a cylinder pulls less than it pushes. With bar and cm², the force comes out in kN divided by 100 (and in kgf, the kN times 102); with psi and in² it comes out in lbf directly. Flow sets the speed, Q = A·v, and the pump power is L/min times bar divided by 600, in kW and before efficiencies. The usual oil is ISO VG 46 or 68 (ISO 3448).

F = p · A   ·   F [kN] = p [bar] · A [cm²]100   ·   F [lbf] = p [psi] · A [in²]   ·   Q = A · v   ·   P [kW] = Q [L/min] · p [bar]600

p: pressure; A: piston area (push) or annular area (pull); v: rod speed; Q: flow. In SI units F (N) = p (Pa) · A (m²). 1 bar = 10 N/cm² = 14.5 psi.

A1 = π·D² / 4A2 = π·(D² − d²) / 4rod dia. dpiston (bore D)Ddp (extend)p (retract)extend F1 = p · A1retract F2 = p · A2F [kN] = p [bar] · A [cm²] / 100     1 kN = 102 kgf = 225 lbf
Hydraulic cylinders: push (extend) force by bore (F = p · A, no friction)
BoreAreaPush in kNPush in kgf
Dia.
mm
A
cm²
100
bar
160
bar
210
bar
250
bar
100
bar
160
bar
210
bar
250
bar
328.048.012.916.920.1820131217222050
4012.5712.620.126.431.41281205026913204
5019.6319.631.441.249.12002320442055006
6331.1731.249.965.577.93179508666757947
8050.2750.380.4105.6125.7512682011076412814
10078.5478.5125.7164.9196.38009128141681920022
125122.72122.7196.3257.7306.812514200222627931285
160201.06201.1321.7422.2502.720503328044305551257
200314.16314.2502.7659.7785.432035512576727480088
Hydraulic cylinders: pull (retract) force with the two main ISO 6020-2 rod sizes for each bore (annular area); from 50 mm up there is also an intermediate rod
BoreRodAnnular areaPull in kNPull in kgf
Dia.
mm
Dia.
mm
A
cm²
100
bar
160
bar
210
bar
250
bar
100
bar
160
bar
210
bar
250
bar
32146.506.510.413.716.3663106113931658
32224.244.26.88.910.64326929081081
401810.0210.016.021.025.11022163521462555
40286.416.410.313.516.0654104613721634
502215.8315.825.333.339.61615258333914036
50369.469.515.119.923.6964154320252411
632825.0125.040.052.562.52551408153576377
634515.2715.324.432.138.21557249132703892
803640.0940.164.184.2100.240886540858410219
805625.6425.641.053.864.12614418354906535
1004562.6462.6100.2131.5156.66387102191341315968
1007040.0640.164.184.1100.140856535857710211
1255698.0998.1156.9206.0245.210002160042100525006
1259059.1059.194.6124.1147.8602796431265615067
16070162.58162.6260.1341.4406.416578265253481441446
160110106.03106.0169.6222.7265.110812172992270527030
20090250.54250.5400.9526.1626.425548408775365163870
200140160.22160.2256.4336.5400.616338261413431040845

100 bar = 1450 psi, 160 bar = 2320 psi, 210 bar = 3046 psi, 250 bar = 3626 psi. 1 kN = 225 lbf.

ISO 15552 pneumatic cylinders at 6 bar (87 psi): push and pull
Bore
mm
Rod dia.
mm
Area
cm²
Push
N
Push
kgf
Annular area
cm²
Pull
N
Pull
kgf
32128.0448349.26.9141542.3
401612.5775476.910.5663364.6
502019.631178120.116.49990100.9
632031.171870190.728.031682171.5
802550.273016307.545.362721277.5
1002578.544712480.573.634418450.5
12532122.727363750.8114.686881701.6
16040201.06120641230.2188.50113101153.3
20040314.16188501922.1301.59180961845.2

A 100 mm pneumatic cylinder at 6 bar (87 psi) pushes 481 kgf; a hydraulic one of the same bore at 210 bar (3000 psi), 16800. That is why air moves and oil clamps.

Hydraulic or pneumatic cylinder: forces

Bore and rod in mm (1 in = 25.4 mm). For a pneumatic cylinder enter the line pressure (6 bar, 87 psi). Forces are theoretical: seal friction takes 5 to 10 %.

Friction, levers and inclined planes

The force to drag something is its weight times the coefficient of friction, which depends on the two materials and on the lubricant. The table values are indicative: steel on dry steel runs from 0.5 to 0.8 with finish and cleanliness, and an oil film brings it down to 0.1. The static coefficient, for starting, is always higher than the kinetic one, for keeping the motion going; that is why a load breaks free with a jerk.

Common coefficients of friction (indicative: they depend on finish, cleanliness and lubricant)
MaterialsStatic μ₀Kinetic μCondition
Steel on steel0.5 to 0.80.4 to 0.6dry
Steel on steel0.1 to 0.160.05 to 0.1lubricated
Steel on gray cast iron0.40.23dry
Steel on gray cast iron0.18 to 0.210.13lubricated
Steel on bronze0.350.3dry
Steel on bronze0.150.1lubricated
Steel on wood0.5 to 0.60.3 to 0.5dry
Rubber on concrete0.6 to 1.00.5 to 0.8dry; wet drops to 0.45
Steel on concrete0.45 to 0.60.4 to 0.5dry (base plates, supports)

Simple machines trade force for travel: the ideal mechanical advantage says how many times the force is multiplied, and the real one is lower because of friction, which on a screw eats most of the effort. The stress a part carries is force over section, σ = F/A, in N/mm² (MPa) or psi, and the safety factor is what the material withstands over what it works at. Beams, columns and buildings are not worked out here: they live in the Engineering Suite. Bolt tightening torque is in chapter 18.

σ = FA   ·   SF = σlimitσworking   ·   Ffriction = μ · N

σ: stress (N/mm² = MPa); F: force (N); A: section (mm²); SF: safety factor (2 to 4 on machine parts, more in lifting gear); μ: coefficient of friction; N: normal force (the weight, on a horizontal support).

Levers, pulleys, inclined plane, wedge and screw: ideal mechanical advantage (MA = load / effort)
MachineMechanical advantageEffort neededNotes
First-class lever (fulcrum in the middle)VM = a / bF = Q · b / aa: effort arm; b: load arm. Crowbar, scissors
Second-class lever (load in the middle)VM = a / b > 1F = Q · b / awheelbarrow, nutcracker
Third-class lever (effort in the middle)VM = a / b < 1F = Q · b / atweezers, the human arm: gains travel, not force
Fixed pulleyVM = 1F = Qonly changes the direction
Movable pulleyVM = 2F = Q / 2the load hangs from two parts of rope
Block and tackle, n parts on the moving blockVM = nF = Q / nreal: F ≈ Q / (n · η), with overall η of 0.8 to 0.95: it drops with each sheave
Windlass (drum r, crank R)VM = R / rF = Q · r / R–
Inclined plane (length L, height h, angle α)VM = L / hF = Q · (sen α + μ · cos α)without friction F = Q · h / L
Wedge (length L, thickness e)VM = L / eF = Q · e / Lideal; friction makes it self-locking
Screw jack (lead p, lever r)VM = 2 · π · r / pF = Q · p / (2 · π · r · η)screw η 0.2 to 0.4: most of the effort goes into friction

Slings: the angle between the legs

Two vertical legs split the load in half. As the angle between them opens, each leg carries more than half, and at 120° each carries the whole load. The factor is 1/cos(θ/2), with θ the included angle: 1.41 at 90°. That is why slings are rated by angle and why a short sling on a wide part is a problem. How lifting lugs and eye bolts are chosen is in the lifting guide.

Q (load)θTTT = Q / (2 · cos(θ/2))tension in each leg:θ = 0°: T = 0.50 · Qθ = 60°: T = 0.58 · Qθ = 90°: T = 0.71 · Qθ = 120°: T = 1.00 · Q
T = Q2 · cos(θ/2)   ·   Qmax = 2 · WLL · cos(θ/2)

T: tension in each leg; Q: weight of the load (same unit); θ: included angle between the two legs; WLL: working load limit of one leg. With unequal legs, the more vertical one carries more.

Two-leg sling: tension in each leg by included angle (T = Q / (2 · cos(θ/2)))
Included angle θAngle to vertical
°
Tension per leg
× Q
Factor
1 / cos(θ/2)
Capacity of the pair
× WLL of one leg
0°00.5001.0002.000
30°150.5181.0351.932
45°230.5411.0821.848
60°300.5771.1551.732
90°450.7071.4141.414
120°601.0002.0001.000

WLL: working load limit of one vertical leg. Included angles over 120° are not used; many standards and makers stop at 90°.

The capacity that counts is the one on the sling tag for the hitch in use (vertical, choker, basket, two legs at a given angle). No tag, no lift.

Frequently asked questions

How do you calculate motor torque from power?

M = 9550 · P / n with P in kW and n in rpm, or T = 5252 · HP / n in lbf·ft. A 5.5 kW motor at 1450 rpm gives 36 N·m; after a 20:1 gearbox at 94 % efficiency, 681 N·m at 72.5 rpm.

How much force does a hydraulic cylinder produce?

Pressure times piston area: F [kN] = p [bar] × A [cm²] / 100, or lbf = psi × in². A 100 mm bore (78.5 cm²) at 210 bar pushes 165 kN, about 16.8 metric tons; retracting with a 70 mm rod, 84 kN.

How many amps does a 10 HP motor draw at 460 V?

About 13 A at full load by the formula (7.5 kW with 0.88 efficiency and 0.84 power factor); the NEC table lists 14 A. At 380 V it is about 15 A. The nameplate value is the one to use for the overload relay.

Sources: ISO 80000-3:2019 and ISO 80000-4:2019 (quantities of space, time and mechanics); IEC 60034-1 (rotating electrical machines: rating and performance); IEC 60072-1 (dimensions and output series of IEC frames 56 to 400) and EN 50347 (power-to-frame assignment); ISO 6020-2 (160 bar compact hydraulic cylinders); ISO 15552 (pneumatic cylinders); ISO 3448 (ISO VG viscosity grades); DIN 15003 (lifting appliances: loads and forces); the lifting guides of the Escuela del Acero. Drive efficiencies and friction coefficients: indicative values from mechanical engineering handbooks and gearbox catalogs.

Keep reading

Chapter 16: V-belts and roller chains

Belt profiles, belt length and chain ratings for the drive downstream of the motor.

Read →

Lifting lugs and eye bolts

How to choose and check them before a lift.

Read →

Handbook index

The twenty-five chapters and the appendices.

See the index →

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