Mechanical Engineering Principles

110 MECHANICAL ENGINEERING PRINCIPLES

Problem 3. The circular hand-wheel of a

valve of diameter 500 mm has a couple

applied to it composed of two forces, each of

250 N. Calculate the torque produced by the

couple.

Torque produced by couple,T=Fd, where force
F=250 N and distance between the forces,
d=500 mm= 0 .5m.
Hence,torque,T=( 250 )( 0. 5 )=125 N m

Now try the following exercise

Exercise 44 Further problems on torque

1. Determine the torque developed when a
   force of 200 N is applied tangentially to
   a spanner at a distance of 350 mm from
   the centre of the nut. [70 N m]


2. During a machining test on a lathe, the
   tangential force on the tool is 150 N. If
   the torque on the lathe spindle is 12 N m,
   determine the diameter of the work-piece.
   [160 mm]

9.2 Work done and power transmitted

by a constant torque

Figure 9.3(a) shows a pulley wheel of radius r
metres attached to a shaft and a forceFNewton’s
applied to the rim at pointP.

r

s

r

P

P

F

(a) F (b)

θ

Figure 9.3

Figure 9.3(b) shows the pulley wheel having turned
through an angleθradians as a result of the forceF

being applied. The force moves through a distance

s, where arc lengths=rθ

Work done=force×distance moved by the force

=F×rθ=FrθNm=FrθJ

However,Fris the torqueT, hence,

work done=Tθjoules

Average power=

work done

time taken

=

Tθ

time taken

for a constant torqueT

However, (angleθ)/(time taken)=angular veloc-

ity,ωrad/s

Hence, power,P=Tωwatts ( 9. 1 )

Angular velocity,ω= 2 πnrad/s wherenis the

speed in rev/s

Hence, power,P=^2 πnTwatts ( 9. 2 )

Sometimes power is in units of horsepower (hp),

where

1 horsepower= 745 .7 watts

i.e. 1hp=^745 .7 watts

Problem 4. A constant force of 150 N is

applied tangentially to a wheel of diameter

140 mm. Determine the work done, in joules,

in 12 revolutions of the wheel.

Torque T=Fr,whereF=150 N and radius

r=

140

2

=70 mm= 0 .070 m.

Hence, torqueT=( 150 )( 0. 070 )= 10 .5Nm.

Work done=Tθjoules, where torque,

T = 10 .5 N m and angular displacement,θ = 12

revolutions= 12 × 2 πrad= 24 πrad.

Hence,work done=( 10. 5 )( 24 π)=792 J

Problem 5. Calculate the torque developed

by a motor whose spindle is rotating at 1000

rev/min and developing a power of 2.50 kW.