We define the displacement vector d to have a magnitude equal to the distance from point A to
point B with a direction the same as the direction of motion (the down-the-ramp direction).
Using our definition of work as the dot product of the force and the displacement, equation 24-4:
W = F ⋅ ∆r
with the gravitational force vector F being the force, and d being the displacement, the work
g
can be written as:
W = F ⋅ d .
g
Using the definition of the dot product we find that:
W = F d cosθ .
g
Replacing the magnitude of the gravitational force with mg we arrive at our final answer:
W = mgd cosθ .
This is the same answer that we got prior to our discussion of the dot product.
In cases in which the force and the displacement vectors are given in i, j, k notation, finding
the work is straightforward.
The Dot Product in Unit Vector Notation
The simple dot product relations among the unit vectors makes it easy to evaluate the dot product
of two vectors expressed in unit vector notation. From what amounts to our definition of the dot
product, equation 24-3:
A ⋅B = AB cosθ
we note that a vector dotted into itself is simply the square of the magnitude of the vector. This
is true because the angle between a vector and itself is 0° and cos 0° is 1.
o
2
A ⋅ A = AAcos0 = A
2
Since the unit vectors all have magnitude 1, any unit vector dotted into itself yields (1) which is
just 1.
i ⋅i = 1, j ⋅ j = 1, and k ⋅ k = 1
Now the angle between any two different Cartesian coordinate axis unit vectors is 90° and the
cos 90° is 0. Thus, the dot product of any Cartesian coordinate axis unit vector into any other
Cartesian coordinate axis unit vector is zero.
1
68