1. Sect. 7.3: Scalar Product of 2 Vectors

2. Scalar Product of 2 Vectors Scalar Product of Two 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

3. AB ≡ AB cosθ W ≡ FΔr Scalar Product of two vectors A & B:
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:
Commutative: AB = BA
Distributive: A(B + C) = AB + AC

4. 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

5. AB = AxBx + AyBy + AzBz (1)
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)

6. AA = (Ax)2 + (Ay)2 + (Az)2
or 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

7. 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 

8. W = (½)kx2 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

9. Applied Force Fapp is equal & opposite to the force
Figure 7.10: A block moves from xi = −xmax to xf = 0 on a frictionless surface as a force Fapp is applied to the block. If the process is carried out very slowly, the applied force is equal in magnitude and opposite in direction to the spring force at all times.
Applied Force Fapp is equal & opposite to the force
Fs exerted by block on spring: Fs = - Fapp = -kx

10. Force Exerted by a Spring on a Block
x > 0, Fs < 0 
Force Fs varies with
block position x
relative to equilibrium
at x = 0. Fs = -kx
spring constant k > 0
x = 0, Fs = 0 
Figure 7.9: The force exerted by a spring on a block varies with the block’s position x relative to the equilibrium position x = 0. (a) When x is positive (stretched spring), the spring force is directed to the left. (b) When x is zero (natural length of the spring), the spring force is zero. (c) When x is negative (compressed spring), the spring force is directed to the right. (d) Graph of Fs versus x for the block-spring system. The work done by the spring force on the block as it moves from −xmax to 0 is the area of the shaded triangle, ½kx 2max.
x < 0, Fs > 0 
Fs (x) vs. x 

11. Work Done to Compress Spring

12. W = (½)kx2

13. 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
Figure 7.11: (Example 7.5) Determining the force constant k of a spring. The elongation d is caused by the attached object, which has a weight mg.