Sect. 7.3: Scalar Product of 2 Vectors
Scalar Product of 2 Vectors
• Pure math discussion for a short time.
• We now know that work done by a constant force is:
W = F||Δr = FΔr cosθ
• Note that both F & Δr are vectors, while W is a scalar.
• W has the mathematical form that mathematicians call
the Scalar Product.
Scalar Product of Two Vectors
• If A & B are vectors, their Scalar Product is defined as:
AB ≡ AB cosθ
Scalar Product of two vectors A & B:
AB ≡ AB cosθ
• So, the work done by a constant force, W =
FΔr cosθ, could be written as
W ≡ FΔr
• Basic math properties of the Scalar Product:
1. Commutative: AB = BA
2. Distributive: A(B + C) = AB + AC
Scalar Product of 2 Vectors Using Components
Scalar Product: AB ≡ AB cosθ
• From the discussion of vector components & unit vectors: A 3
dimensional vector V can be written as
V = Vxi + Vyj + Vzk
i,j,k are unit vectors along the x,y,z axes, 
Vx,Vy,Vz are the x,y,z components of V
Unit vectors i,j,k are dimensionless & have length 1
• From scalar product definition & because the angles between
i,j,k are all 90 [cos(90°) = 0, cos(0°) = 1] we can easily show:
ii = jj = kk = 1
ij= ik = jk = 0
• For two 3 d vectors A & B using this notation:
A = Axi + Ayj + Azk, B = Bxi + Byj + Bzk
• Consider the scalar product of A & B in this form:
AB = (Axi + Ayj + Azk)(Bxi + Byj + Bzk)
• Rewriting this (long form, 9 terms!):
AB =
(AxBx)(ii) + (AxBy)(ij) + (AxBz)(ik) +
(AyBx)(ji) + (AyBy)(jj) + (AyBz)(jk) +
(AzBx)(ki) + (AzBy)(kj) + (AzBz)(kk)
• Using ii = jj = kk = 1, ij= ik = jk = 0 gives
AB = AxBx + AyBy + AzBz
(1)
This form will be convenient sometimes!
• Also, consider the case where A = B. Then, (1) becomes:
AA = AxAx + AyAy + AzAz
or
AA = (Ax)2 + (Ay)2 + (Az)2
(2)
or
AA = (Ax)2 + (Ay)2 + (Az)2
AA = |A|2 = square of length of A
 The scalar product of a vector with itself is the
square of the length of that vector.
• Examples 7.2 & 7.3
Sect. 7.4: Work Done by a Variable Force
• See text for details. Requires that you know simple integral
calculus.
• In one dimension, for F = F(x), the bottom line is that the work
is the integral of the F vs. x curve:
W = ∫ F(x) dx (limits xi to xf)
• For those who don’t understand
integrals, this is THE AREA
under the F vs. x curve

• Consider an ideal spring:
Characterized by a spring constant k.
A measure of how “stiff” the spring
is. Hooke’s “Law” restoring force
Fs = -kx (Fs >0, x <0; Fs <0, x >0)
Work done by person:
W = (½)kx2
Applied Force Fapp is equal & opposite to the force
Fs exerted by block on spring: Fs = - Fapp = -kx
Force Exerted by a Spring on a Block
Force Fs varies with
block position x
relative to equilibrium
at x = 0. Fs = -kx
spring constant k > 0
x > 0, Fs < 0 
x = 0, Fs = 0 
x < 0, Fs > 0 
Fs (x) vs. x 
Work Done to Compress Spring
W = (½)kx2
Example 7.5: Measuring k for a Spring
Hang spring vertically.
Attach object of mass
m to lower end. Spring
Stretches a distance d.
At equilibrium, N’s 2nd
Law says ∑Fy = 0.
So, mg – kd = 0, mg = kd
Know m, measure d,
 k = (mg/d)
Example:
d = 2.0 cm, m = 0.55 kg
 k = 270 N/m