082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

244 Chapter 7 Vectors
Engineeringapplication7.7
Workdonebyaforce
Ifaforceisappliedtoanobjectandtheobjectmoves,thenworkisdonebytheforce.
It is possible to use vectors to calculate the work done. Suppose a constant force F
is applied and as a consequence the object moves from A to B, represented by the
displacement vectors, asshown inFigure 7.26.
Wecanresolvetheforceintotwoperpendicularcomponents,oneparalleltosand
oneperpendiculartos.Theworkdonebyeachcomponentisequaltotheproductof
itsmagnitudeandthedistancemovedinitsdirection.Thecomponentperpendicular
to s will not do any work because there is no movement in this direction. For the
component alongs, that is |F|cos θ, we find
work done = |F|cos θ|s|
From the definition of the scalar product we see that the r.h.s. of this expression is
the scalar product ofFands.
F cos u
F
F sin u
u
A
AB = s
B
Figure7.26
ThecomponentofFinthedirectionofs
isFcosθ.
The work done by a constant force F which moves an object along the vector s is
equal tothe scalar productF·s.
Example7.17 A force F = 3i + 2j − k is applied to an object which moves through a displacement
s = 2i +2j +k. Find the work done by the force.
Solution The work done isequal to
F·s=(3i+2j−k)·(2i+2j+k)
=6+4−1
= 9 joules
Engineeringapplication7.8
Movementofachargedparticleinanelectricfield
Figure7.27showstwochargedplatessituatedinavacuum.PlateAhasanexcessof
positivecharge,whileplateBhasanexcessofnegativecharge.Suchanarrangement
gives risetoan electric field.An electricfield isan example of avector field.