Sect. 7.3: Scalar Product of 2 Vectors

1. Sect. 7.3: Scalar Product of 2 Vectors

2. 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, theirScalar Productis defined as: AB ≡ AB cosθ

3. 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: • 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,zaxes,  Vx,Vy,Vzare the x,y,zcomponents 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. 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 = 0gives 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+ AzAzor 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 xito xf) • For those who don’t understand integrals, this is THE AREA under the F vs. x curve 

8. 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 Fappis 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 Fsvaries with block position x relative to equilibrium at x = 0. Fs = -kx spring constant k > 0 x = 0, Fs = 0  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