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Vector Sum

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Vector Sum


1. State the problem: Which diagram satisfies the vector equation $\vec{C}=\vec{A}+\vec{B}$? 1. Explain the criterion: The sum $\vec{A}+\vec{B}$ is formed by placing $\vec{B}$ head-to-tail with $\vec{A}$ and the resultant is the vector from the tail of $\vec{A}$ to the head of $\vec{B}$. 1. Conclusion: Diagram (a) satisfies $\vec{C}=\vec{A}+\vec{B}$ because it places $\vec{A}$ and $\vec{B}$ head-to-tail and $\vec{C}$ connects the tail of $\vec{A}$ to the head of $\vec{B}$. 2. State the problem: Given $A=10\ \text{cm}$ at $\alpha=53^\circ$, $B=7\ \text{cm}$ at $\beta=120^\circ$, and $C=8\ \text{cm}$ at $\gamma=210^\circ$, find (a) $\vec{R}=\vec{A}+\vec{B}$, (b) $\vec{P}=\vec{A}-\vec{B}$, and (c) $\vec{Q}=\vec{A}-2\vec{B}+\vec{C}$ using a component method. 2. Step: Compute components of each vector. $$A_x=A\cos\alpha=10\cos 53^\circ=10(0.6)=6\ \text{cm}$$ $$A_y=A\sin\alpha=10\sin 53^\circ=10(0.8)=8\ \text{cm}$$ $$B_x=B\cos\beta=7\cos 120^\circ=7(-0.5)=-3.5\ \text{cm}$$ $$B_y=B\sin\beta=7\sin 120^\circ=7(0.8660254)\approx 6.0621778\ \text{cm}$$ $$C_x=C\cos\gamma=8\cos 210^\circ=8(-0.8660254)\approx -6.9282032\ \text{cm}$$ $$C_y=C\sin\gamma=8\sin 210^\circ=8(-0.5)=-4\ \text{cm}$$ 2(a). Compute $\vec{R}=\vec{A}+\vec{B}$ components and magnitude and direction. $$R_x=A_x+B_x=6+(-3.5)=2.5\ \text{cm}$$ $$R_y=A_y+B_y=8+6.0621778=14.0621778\ \text{cm}$$ $$|\vec{R}|=\sqrt{R_x^2+R_y^2}=\sqrt{2.5^2+14.0621778^2}\approx 14.285\ \text{cm}$$ $$\theta_R=\arctan\left(\frac{R_y}{R_x}\right)\approx 80.0^\circ\ \text{(from +x)}$$ 2(b). Compute $\vec{P}=\vec{A}-\vec{B}$ components and magnitude and direction. $$P_x=A_x-B_x=6-(-3.5)=9.5\ \text{cm}$$ $$P_y=A_y-B_y=8-6.0621778=1.9378222\ \text{cm}$$ $$|\vec{P}|=\sqrt{P_x^2+P_y^2}=\sqrt{9.5^2+1.9378222^2}\approx 9.695\ \text{cm}$$ $$\theta_P=\arctan\left(\frac{P_y}{P_x}\right)\approx 11.5^\circ\ \text{(from +x)}$$ 2(c). Compute $\vec{Q}=\vec{A}-2\vec{B}+\vec{C}$ components and magnitude and direction. $$(-2\vec{B})_x=-2B_x=-2(-3.5)=7.0\ \text{cm}$$ $$(-2\vec{B})_y=-2B_y=-2(6.0621778)\approx -12.1243556\ \text{cm}$$ $$Q_x=A_x+(-2B)_x+C_x=6+7.0+(-6.9282032)\approx 6.0717968\ \text{cm}$$ $$Q_y=A_y+(-2B)_y+C_y=8+(-12.1243556)+(-4)\approx -8.1243556\ \text{cm}$$ $$|\vec{Q}|=\sqrt{Q_x^2+Q_y^2}=\sqrt{6.0717968^2+(-8.1243556)^2}\approx 10.144\ \text{cm}$$ $$\theta_Q=\arctan\left(\frac{Q_y}{Q_x}\right)\approx -53.3^\circ=306.7^\circ\ \text{(from +x)}$$ 2(d). Question: Is the magnitude of $\vec{C}$ equal to the sum of magnitudes $A+B$? Answer: No in general the magnitude of a resultant is not the arithmetic sum of magnitudes, because directions matter. 2(d). Explanation: By the triangle inequality $|\vec{A}+\vec{B}|\le |\vec{A}|+|\vec{B}|$ with equality only when $\vec{A}$ and $\vec{B}$ are parallel and in the same direction. 2(e). Question: Is the magnitude of $\vec{C}$ equal to the difference of magnitudes $A-B$? Answer: No in general the magnitude is not the difference, and $|\vec{A}+\vec{B}|$ equals $||\vec{A}|-|\vec{B}||$ only when $\vec{A}$ and $\vec{B}$ are collinear and in opposite directions. 3. State the problem: An insect moves from point $\,(12,8)\,$ to $\,(7,4)\,$ in the coordinate system with origin at the lower-left corner; find the displacement vector. 3. Compute displacement: The displacement is $\vec{\Delta r}=\vec{r}_{\text{final}}-\vec{r}_{\text{initial}}$. $$\vec{\Delta r}=(7-12,\;4-8)=(-5,\;-4)\ \text{cm}$$ 3. Magnitude and direction: $$|\vec{\Delta r}|=\sqrt{(-5)^2+(-4)^2}=\sqrt{25+16}=\sqrt{41}\approx 6.403\ \text{cm}$$ $$\theta_{\Delta r}=\arctan\left(\frac{-4}{-5}\right)\approx 218.66^\circ\ \text{(from +x, measured counterclockwise)}$$ 4. State the problem: What is the minimum number of equal-magnitude vectors required to produce a zero resultant? 4. Answer: Two equal-magnitude vectors can produce a zero resultant if they are equal in magnitude and opposite in direction, so the absolute minimum is 2. 4. Clarification: If the vectors are required to be noncollinear (no two along the same line), then three vectors of equal magnitude arranged at 120 degree intervals will produce zero resultant.