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PRC Mechanical Engineer practice questions

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Question 1Machine Design, Materials and Shop Practice

A rectangular parallel key is used to connect a pulley to a shaft. The key principally carries which type of stress?

  1. Torsion about the key's own centroidal axis
  2. Shear on the plane between the shaft and the hub Correct answer.
  3. Bending about the key's long axis
  4. Direct tension along the length of the key

Explanation

The key fills matching keyways in the shaft and hub, so torque is transferred across the shaft-hub interface plane and the key is sheared on that plane. Its side faces also carry bearing (crushing) pressure, and keys are normally sized from the shear strength of the key material using the shaft torque and radius.

Question 2Machine Design, Materials and Shop Practice

Grey cast iron is widely used for machine-tool beds and frames mainly because of its:

  1. High internal damping capacity Correct answer.
  2. High ductility at room temperature
  3. High tensile strength
  4. Low compressive strength

Explanation

The flake graphite in grey cast iron absorbs vibration energy, giving it excellent damping capacity, which keeps chatter and vibration low in machine tools. It is also cheap, castable into complex shapes and strong in compression, though weak and brittle in tension.

Question 3Machine Design, Materials and Shop Practice

Which hardness test uses a hardened 10 mm steel ball pressed under a standard load and is well suited to coarse-grained cast irons?

  1. Rockwell C
  2. Brinell Correct answer.
  3. Shore scleroscope
  4. Vickers

Explanation

The Brinell test presses a 10 mm hardened steel (or carbide) ball into the surface and relates hardness to the diameter of the impression, so the large indenter averages out the structure. This makes it suitable for coarse-grained materials such as cast iron and for soft or inhomogeneous metals.

Question 4Machine Design, Materials and Shop Practice

Which welded joint is made by bringing two plate edges together in the same plane and gives the highest strength under tensile loading?

  1. Butt joint Correct answer.
  2. Tee joint
  3. Lap joint
  4. Corner joint

Explanation

In a butt joint the two plates lie in the same plane and the weld joins their edges, so the load path stays straight through the section and full-penetration butt welds can approach the strength of the parent plate. Lap, tee and corner joints introduce eccentricity or a change of direction in the load path, which lowers joint efficiency.

Question 5Machine Design, Materials and Shop Practice

In a belt drive, the speed of the driven pulley divided by the speed of the driver pulley equals:

  1. Product of the two pulley diameters
  2. Driver diameter divided by driven diameter Correct answer.
  3. Driven diameter divided by driver diameter
  4. Square of the diameter ratio

Explanation

For negligible slip, the belt surface speed is the same on both pulleys, so n1·D1 = n2·D2 and therefore n2/n1 = D1/D2. A large driven pulley thus turns more slowly than a small driver.

Question 6Machine Design, Materials and Shop Practice

In surface-texture measurement, the parameter Ra denotes the:

  1. Root-mean-square slope of the roughness profile
  2. Maximum peak-to-valley height of the sampled profile
  3. Arithmetic average roughness from the mean line Correct answer.
  4. Average spacing between successive profile peaks

Explanation

Ra is the arithmetic mean of the absolute deviations of the roughness profile from its mean line over the sampling length. It is the most widely specified roughness parameter because it is stable and easy to measure, although it does not describe the shape of the peaks.

Question 7Machine Design, Materials and Shop Practice

Tempering of a hardened steel component is carried out mainly to:

  1. Relieve internal stresses and restore toughness Correct answer.
  2. Refine the grain size of the prior austenite
  3. Increase the surface hardness above the as-quenched value
  4. Produce a fully austenitic structure

Explanation

Tempering reheats the as-quenched martensitic steel to a temperature below the lower critical temperature, allowing carbon to precipitate and residual stresses to relax. Hardness and strength drop somewhat while toughness and ductility rise, which is the whole point of the treatment.

Question 8Machine Design, Materials and Shop Practice

The factor of safety of a machine member is defined as the ratio of:

  1. Failure strength to allowable stress Correct answer.
  2. Working load to proof load
  3. Strain at fracture to strain at yield
  4. Working stress to yield strength

Explanation

Factor of safety compares the strength at which the member is considered to fail (yield or ultimate, depending on the design criterion) with the stress permitted in service. An FoS greater than one means the working stress is below the failure strength.

Question 9Machine Design, Materials and Shop Practice

For wrought steels with a tensile strength below about 1400 MPa, the rotating-beam endurance limit is approximately:

  1. 0.9 times the tensile strength
  2. 0.1 times the tensile strength
  3. 0.5 times the tensile strength Correct answer.
  4. Equal to the tensile strength

Explanation

Fatigue data for steels show that the fully reversed bending endurance limit is roughly half the ultimate tensile strength up to about 1400 MPa, beyond which the ratio falls. This rule of thumb is the usual starting point before applying surface, size and reliability factors.

Question 10Machine Design, Materials and Shop Practice

The main purpose of preloading a bolted joint is to:

  1. Permit joint separation at low external load
  2. Raise the thread friction coefficient in service
  3. Eliminate the need for a proof load check
  4. Clamp members so the bolt sees less cyclic load Correct answer.

Explanation

Preload compresses the clamped members so that an externally applied load mainly relieves this clamping rather than stretching the bolt. The bolt therefore experiences only a fraction of the external cyclic load, which greatly improves its fatigue life and resists self-loosening.

Question 11Machine Design, Materials and Shop Practice

In the hole-basis system of limits and fits:

  1. Both hole and shaft carry bilateral tolerances about the basic size
  2. The shaft size is held constant and the hole is varied
  3. The hole's lower deviation is zero and shaft deviations are varied Correct answer.
  4. Only interference fits can be specified

Explanation

In the hole-basis system the hole is made to a basic size with fundamental deviation H, so its lower deviation is zero and its tolerance lies above basic size. Different classes of fit are then obtained by changing the shaft deviation, which is economical because holes can be produced with fixed reamers and gauges.

Question 12Machine Design, Materials and Shop Practice

The module of a spur gear is defined as the:

  1. Circular pitch multiplied by π
  2. Number of teeth divided by pitch diameter
  3. Pitch diameter divided by number of teeth Correct answer.
  4. Addendum divided by 2.25

Explanation

Module m = d/z, where d is the pitch-circle diameter and z the number of teeth, expressed in millimetres. It is the standard measure of tooth size and is related to circular pitch by p = πm.

Question 13Machine Design, Materials and Shop Practice

A ball bearing carries a radial load P at constant speed. If the load is doubled, its L10 rating life becomes:

  1. One-half of its original value
  2. Twice its original value
  3. One-quarter of its original value
  4. One-eighth of its original value Correct answer.

Explanation

For ball bearings the basic rating life varies as (C/P)^3, where C is the basic dynamic load rating. Doubling P therefore reduces L10 by a factor of 2^3 = 8.

Question 14Machine Design, Materials and Shop Practice

The spring index of a helical compression spring is the ratio of:

  1. Free length to wire diameter
  2. Wire diameter to mean coil diameter
  3. Mean coil diameter to wire diameter Correct answer.
  4. Solid length to pitch

Explanation

Spring index C = D/d, with D the mean coil diameter and d the wire diameter. It controls stress concentration and manufacturability, with values of roughly 4 to 12 being practical; small indices give high curvature stress.

Question 15Machine Design, Materials and Shop Practice

A 50 mm diameter bar is turned at a cutting speed of 100 m/min. The spindle speed is closest to:

  1. 477 rpm
  2. 637 rpm Correct answer.
  3. 318 rpm
  4. 1274 rpm

Explanation

Spindle speed follows N = 1000V/(πD), so N = 1000 × 100/(π × 50) ≈ 637 rpm. The factor of 1000 converts the diameter from millimetres to metres.

Question 16Machine Design, Materials and Shop Practice

For an M10 × 1.5 coarse thread, the recommended tapping drill diameter is:

  1. 10.0 mm
  2. 8.5 mm Correct answer.
  3. 7.5 mm
  4. 9.5 mm

Explanation

For metric threads the tapping drill size is approximately the major diameter minus the pitch, giving 10 − 1.5 = 8.5 mm. This leaves about 75 per cent thread engagement, which gives good strength without overloading the tap.

Question 17Machine Design, Materials and Shop Practice

The theoretical stress concentration factor Kt for a notched member is defined as the:

  1. Maximum local stress divided by nominal stress Correct answer.
  2. Fatigue limit divided by static strength
  3. Yield strength divided by nominal stress
  4. Nominal stress divided by maximum local stress

Explanation

Kt compares the peak elastic stress at a geometric discontinuity with the nominal stress computed on the net or gross section. Because it is derived from elastic theory, it depends on geometry alone and is then modified by notch sensitivity for fatigue work.

Question 18Machine Design, Materials and Shop Practice

For a single-plate friction clutch, the torque transmitted is proportional to the:

  1. Clamping force times friction coefficient times mean radius Correct answer.
  2. Axial clamping force multiplied by the outer radius of the plate
  3. Angular velocity times axial force
  4. Coefficient of friction times plate thickness

Explanation

Uniform-wear or uniform-pressure theory gives T = μ W r_m for one pair of friction faces, where W is the axial clamping force and r_m the mean radius. Adding more plates multiplies the capacity by the number of effective friction faces.

Question 19Machine Design, Materials and Shop Practice

Creep of a material is:

  1. Reversible strain produced by cyclic loading
  2. Elastic springback observed immediately after unloading at room temperature
  3. Time-dependent deformation under sustained elevated-temperature stress Correct answer.
  4. Sudden brittle fracture under impact at room temperature

Explanation

Creep is progressive, permanent deformation that occurs while a component is held under a steady load at a temperature high relative to its melting point. It is described by primary, secondary and tertiary stages, and design often limits secondary-stage creep rate.

Question 20Machine Design, Materials and Shop Practice

For a solid circular shaft under combined steady bending moment M and torque T, the maximum shear stress theory gives an equivalent twisting moment equal to:

  1. Half of [M + square root of (M squared + T squared)]
  2. Square root of (M squared + T squared) Correct answer.
  3. Square root of (M squared minus T squared)
  4. M + T

Explanation

The principal shear stress under combined bending and torsion is proportional to √(M² + T²), so the equivalent twisting moment is Te = √(M² + T²). The shaft is then sized from τ = 16Te/(πd³).

Question 21Machine Design, Materials and Shop Practice

A cutting tool obeys Taylor's equation VT^0.25 = C. If tool life is 60 min at 60 m/min, the cutting speed for a tool life of 30 min is closest to:

  1. 85 m/min
  2. 71 m/min Correct answer.
  3. 120 m/min
  4. 51 m/min

Explanation

Taylor's equation gives V2 = V1 (T1/T2)^n = 60 × (60/30)^0.25 = 60 × 1.189 ≈ 71 m/min. Because the exponent is small, a large change in tool life causes only a modest change in speed.

Question 22Machine Design, Materials and Shop Practice

The Jominy end-quench test is used to determine the:

  1. Residual austenite content after tempering
  2. Fatigue limit of a welded joint
  3. Wear resistance of a carburised case
  4. Hardenability of a steel grade Correct answer.

Explanation

In the Jominy test a standard bar is austenitised and quenched by a water jet at one end; hardness is then measured along the bar. The resulting hardness-versus-distance curve expresses hardenability, that is, how deep a section can be hardened, which depends mainly on the alloy content.

Question 23Machine Design, Materials and Shop Practice

Two plates of thickness t are joined by rivets of diameter d pitched at p in a lap joint. Failure occurs by tearing of the plate across the rivet line. The joint efficiency is:

  1. d/p
  2. (p − d)/p Correct answer.
  3. p/(p − d)
  4. (p − d)/d

Explanation

The weakest plate section between rivet holes has area (p − d)t while the solid plate has area pt, so the tearing efficiency is (p − d)/p. The overall joint efficiency is the lowest of the tearing, shearing and crushing efficiencies.

Question 24Machine Design, Materials and Shop Practice

A steel hub is shrink-fitted onto a solid steel shaft. Within the elastic range, doubling the diametral interference, with the same materials and geometry, will:

  1. Double the contact pressure Correct answer.
  2. Leave the contact pressure unchanged
  3. Quadruple the contact pressure
  4. Halve the contact pressure

Explanation

For a linear-elastic shrink fit, the radial interference is proportional to the interface pressure through the Lamé coefficients of the two members, so pressure scales directly with interference. Stresses in the hub also double, which is why the elastic limit of the hub bore usually limits the maximum interference.

Question 25Machine Design, Materials and Shop Practice

In the modified Goodman criterion for fatigue design, with alternating stress sa, mean stress sm, endurance limit Se and ultimate strength Su, failure is predicted when:

  1. sa/Se + sm/Su = 1 Correct answer.
  2. sa/Su + sm/Se = 1
  3. sm/Se + sa/Sy = 1
  4. sa/Se − sm/Sy = 1

Explanation

The Goodman line joins the endurance limit on the alternating-stress axis to the ultimate strength on the mean-stress axis, so the intercepts are Se and Su respectively. A design point is safe when sa/Se + sm/Su is less than one, and an equivalent fully reversed stress can be obtained by multiplying both sides by Se.

Question 26Machine Design, Materials and Shop Practice

In grey cast iron, the carbon that is not combined with iron is present as:

  1. temper carbon nodules
  2. spheroidal graphite
  3. flake graphite Correct answer.
  4. cementite lamellae

Explanation

Grey cast iron solidifies slowly enough for most of its carbon to precipitate as flake graphite in a ferrite/pearlite matrix. These flakes act as internal dampers and give the material excellent vibration absorption and machinability, but they also lower its tensile strength.

Question 27Machine Design, Materials and Shop Practice

The standard Brinell hardness test presses into the specimen:

  1. a 1 mm ball under a light load
  2. a 10 mm ball under a specified load Correct answer.
  3. a 136° diamond pyramid under 5 kgf
  4. a 120° diamond cone under 150 kgf

Explanation

Brinell hardness is defined by the indentation diameter left by a hardened steel or carbide ball, normally 10 mm diameter, loaded for a fixed dwell time. The hardness number is the load divided by the curved surface area of the impression.

Question 28Machine Design, Materials and Shop Practice

The main purpose of normalizing a forged steel part is to:

  1. soften the part until it is nearly pure ferrite
  2. refine the grain and relieve internal stresses Correct answer.
  3. raise the carbon content of the surface layer
  4. harden the surface by diffusing carbon into it

Explanation

Normalizing heats the steel above its upper critical temperature and allows it to cool in still air, which produces a fine, uniform grain structure. The result is removal of forging stresses, improved machinability and more consistent mechanical properties.

Question 29Machine Design, Materials and Shop Practice

Cemented carbide tools allow much higher cutting speeds than high-speed steel chiefly because carbide has:

  1. a lower cost per cutting edge
  2. a lower coefficient of friction
  3. better hot hardness Correct answer.
  4. greater toughness under impact

Explanation

At the temperatures generated by high cutting speeds, HSS softens rapidly while cemented carbide retains most of its hardness. This hot hardness is what permits carbide to cut at several times the speed used with HSS.

Question 30Machine Design, Materials and Shop Practice

The gib head provided on a sunk key is used to:

  1. increase the torque capacity
  2. prevent the key from rotating
  3. locate the hub axially on the shaft
  4. withdraw the key from the shaft Correct answer.

Explanation

A gib head is a projecting lug at one end of the key that gives a grip for pulling the key out of its keyway during dismantling. It adds nothing to the load-carrying capacity of the key.

Question 31Machine Design, Materials and Shop Practice

In a sand mould, a riser is provided mainly to:

  1. support the sand core in position
  2. vent steam and gases from the mould cavity
  3. feed metal to offset solidification shrinkage Correct answer.
  4. reduce turbulence as metal enters the mould

Explanation

A riser is a reservoir of liquid metal that continues to feed the casting as it contracts during solidification. It is designed so that it solidifies after the casting, preventing shrinkage cavities in the part.

Question 32Machine Design, Materials and Shop Practice

A V-belt can transmit more power than a flat belt of the same width because:

  1. wedge action in the groove raises friction Correct answer.
  2. it runs with much lower centrifugal tension
  3. it is made of stronger rubberised cord
  4. it needs no initial tension in the belt

Explanation

The tapered sides of a V-belt wedge into the pulley groove, so the normal force on the contact faces is greater than the belt tension alone. This amplified normal force gives a much higher effective friction coefficient than a flat belt on a plain rim.

Question 33Machine Design, Materials and Shop Practice

Parallel misalignment between two shafts is best accommodated by:

  1. a sleeve (muff) coupling
  2. a rigid flange coupling
  3. a universal joint
  4. an Oldham coupling Correct answer.

Explanation

An Oldham coupling uses two hubs with slots and a floating centre disc whose tongues slide in the slots, which allows the shafts to be offset in parallel while still transmitting torque. Rigid couplings demand near-perfect alignment.

Question 34Machine Design, Materials and Shop Practice

A 50 mm diameter shaft transmits 500 N·m of torque through a 12 mm wide square key. If the allowable shear stress in the key is 40 MPa, the minimum key length is closest to:

  1. 42 mm Correct answer.
  2. 84 mm
  3. 21 mm
  4. 63 mm

Explanation

The tangential force on the key is T/r = 500/0.025 = 20 000 N, and the shear area is the width times the length. Setting 20 000/(12L) = 40 MPa gives L = 41.7 mm, so a 42 mm key is the minimum.

Question 35Machine Design, Materials and Shop Practice

Two parallel fillet welds, each of leg size 8 mm and length 100 mm, carry a load of 90 kN. The average shear stress on the weld throat is closest to:

  1. 80 MPa Correct answer.
  2. 113 MPa
  3. 60 MPa
  4. 100 MPa

Explanation

The throat thickness of a 45° fillet is 0.707 times the leg, giving 5.66 mm. The total throat area is 2 × 5.66 × 100 = 1131 mm², so the stress is 90 000/1131 ≈ 80 MPa.

Question 36Machine Design, Materials and Shop Practice

For 20° full-depth involute gears, the minimum number of teeth on the pinion to avoid undercutting is:

  1. 14 teeth
  2. 20 teeth
  3. 12 teeth
  4. 17 teeth Correct answer.

Explanation

With a standard 20° pressure angle and full-depth teeth, the limiting case of a rack cutter gives a minimum of 17 teeth on the pinion. Fewer teeth cause the cutter to undercut the involute near the base circle.

Question 37Machine Design, Materials and Shop Practice

A deep-groove ball bearing has a basic dynamic load rating of 30 kN and carries an equivalent load of 5 kN. Its rating life in million revolutions is:

  1. 6
  2. 36
  3. 216 Correct answer.
  4. 150

Explanation

For ball bearings the life in million revolutions is (C/P)^3, where C is the dynamic rating and P the equivalent load. Here (30/5)^3 = 6^3 = 216 million revolutions.

Question 38Machine Design, Materials and Shop Practice

The spring rate of a helical compression spring is inversely proportional to:

  1. the modulus of rigidity of the wire
  2. the number of active coils Correct answer.
  3. the fourth power of the wire diameter
  4. the cube of the mean coil diameter

Explanation

From k = Gd⁴/(8D³N), the rate varies inversely with the number of active coils only. Adding coils makes the spring softer, while thicker wire or a smaller mean coil diameter makes it stiffer.

Question 39Machine Design, Materials and Shop Practice

A power screw of mean diameter dm and lead l is self-locking when the coefficient of friction satisfies:

  1. μ ≥ l/(π dm) Correct answer.
  2. μ ≥ l/(2π dm)
  3. μ ≥ π dm/l
  4. μ ≤ l/(π dm)

Explanation

Self-locking requires the friction angle to be at least as large as the helix angle, that is μ ≥ tan λ. Since tan λ = l/(π dm), the condition becomes μ ≥ l/(π dm).

Question 40Machine Design, Materials and Shop Practice

Carburising is applied to a low-carbon steel to:

  1. produce a hard wear-resistant case over a tough core Correct answer.
  2. raise the core strength while leaving a soft skin
  3. remove internal stresses before machining
  4. convert the surface layer into flake graphite

Explanation

Carburising diffuses carbon into the surface of a low-carbon steel, and a subsequent quench hardens only that carbon-rich skin. The result is a hard, wear-resistant case supported by a soft, shock-resistant core.

Question 41Machine Design, Materials and Shop Practice

In orthogonal cutting the chip thickness ratio is 0.5 and the rake angle is 10°. The shear plane angle is closest to:

  1. 28° Correct answer.
  2. 45°
  3. 22°
  4. 25°

Explanation

Using tan φ = r cos α/(1 − r sin α) with r = 0.5 and α = 10° gives tan φ = 0.4924/0.9132 = 0.539, so φ ≈ 28°.

Question 42Machine Design, Materials and Shop Practice

A 20 mm diameter hole is drilled through a 60 mm thick plate at 500 rev/min and a feed of 0.2 mm/rev. The drilling time is closest to:

  1. 72 s
  2. 18 s
  3. 120 s
  4. 36 s Correct answer.

Explanation

The feed rate is 0.2 × 500 = 100 mm/min, so the time for 60 mm of travel is 60/100 = 0.6 min, or 36 s.

Question 43Machine Design, Materials and Shop Practice

A column whose slenderness ratio is below the transition value should be designed using:

  1. Johnson's parabolic formula Correct answer.
  2. the Rankine-Gordon constant
  3. the secant formula with zero eccentricity
  4. the Euler formula

Explanation

Euler's formula applies only to long, slender columns that fail by elastic buckling. For intermediate columns that yield before buckling elastically, the Johnson parabolic formula gives the allowable stress.

Question 44Machine Design, Materials and Shop Practice

Shot peening improves the fatigue life of a steel part by:

  1. increasing surface roughness for oil retention
  2. inducing compressive residual stress in the surface Correct answer.
  3. raising the surface hardness to the core hardness
  4. removing residual stresses from the surface

Explanation

Bombarding the surface with small hard shot plastically stretches the outer layer, which is restrained by the unaffected core and therefore ends up in compression. Cracks initiate less readily in a compressive surface layer, so fatigue life increases.

Question 45Machine Design, Materials and Shop Practice

Dye penetrant inspection is used to reveal:

  1. variations in hardness near a weld
  2. internal porosity deep inside a casting
  3. sub-surface cracks in thick sections
  4. surface-breaking defects in non-porous parts Correct answer.

Explanation

The penetrant is drawn by capillary action into defects that reach the surface and is then developed out as a visible indication. It therefore detects only surface-breaking flaws and only in non-porous materials.

Question 46Machine Design, Materials and Shop Practice

For a helical spring with spring index 8, the Wahl factor Kw = (4C − 1)/(4C − 4) + 0.615/C is closest to:

  1. 1.18 Correct answer.
  2. 1.08
  3. 1.45
  4. 1.25

Explanation

Substituting C = 8 gives 31/28 = 1.107 for the curvature term plus 0.615/8 = 0.077 for the direct shear term. The sum is about 1.18.

Question 47Machine Design, Materials and Shop Practice

A notched steel bar has a theoretical stress concentration factor Kt of 3.0 and a notch sensitivity q of 0.80. The fatigue stress concentration factor Kf is:

  1. 2.40
  2. 3.00
  3. 2.60 Correct answer.
  4. 1.80

Explanation

Notch sensitivity is defined by q = (Kf − 1)/(Kt − 1), so Kf = 1 + q(Kt − 1) = 1 + 0.8 × 2 = 2.6. This value, not Kt, is used in the fatigue design equation.

Question 48Machine Design, Materials and Shop Practice

In a bolted joint the bolt stiffness is kb and the stiffness of the clamped members is km. The fraction of an external tensile load that is added to the bolt is:

  1. km/(kb + km)
  2. kb/(kb + km) Correct answer.
  3. kb/(kb − km)
  4. (kb + km)/kb

Explanation

The bolt and the clamped members share the external load in proportion to their stiffnesses, so the bolt takes the fraction kb/(kb + km). A stiff bolt and soft members therefore increase the fluctuating load on the bolt.

Question 49Machine Design, Materials and Shop Practice

A sphere and a cube of equal volume are cast in the same sand mould. By Chvorinov's rule, the ratio of the solidification time of the sphere to that of the cube is closest to:

  1. 0.81
  2. 1.65
  3. 1.24 Correct answer.
  4. 1.00

Explanation

Solidification time varies as (V/A)². For equal volumes the sphere has modulus r/3 while the cube has modulus a/6, and with a³ = (4/3)πr³ this gives a/6 = 0.269r. The ratio is therefore (0.333/0.269)² ≈ 1.24, so the sphere feeds for longer.

Question 50Machine Design, Materials and Shop Practice

A spur gear of 100 mm pitch diameter transmits 10 kW at 1000 rev/min. The tangential tooth load is closest to:

  1. 3000 N
  2. 1910 N Correct answer.
  3. 955 N
  4. 6000 N

Explanation

The pitch-line velocity is π × 0.1 × 1000/60 = 5.24 m/s. Dividing the transmitted power by this velocity gives Ft = 10 000/5.24 ≈ 1910 N.