Question

a)Find F=A+B where A=(150,0 degrees) and (300, 60 degrees). Show work graphically

b) analytically (using law of cosines)

c) analytically (using component vectors)

Answer 1

for b question using Law of Cosines... the included angle between vec A and vec B is \(180^\circ-60^\circ=120^\circ\)... and to find magnitude of vec F\[|\vec F|=\sqrt{150^2+300^2-2(150)(300)\cos120^\circ}=396.863\]

Answer 2

can use Sine Law to solve the angle between vec A and vec F, opposite to vec B...\[\frac{\sin\theta}{300}=\frac{\sin120^\circ}{396.863}\]\[\theta=\sin^{-1}\left[\frac{300\sin120^\circ}{396.863}\right]=40.89^\circ\]therefore \[\vec F=396.863\angle40.89^\circ\]

Answer 3

for c... \[\vec F=\vec{F_x}+\vec{F_y}\]where\[\vec{F_x}=|\vec F|\cos\theta=\vec {A_x}+\vec{B_x}~~~~(1)\]and\[\vec{F_y}=|\vec F|\sin\theta=\vec{A_y}+\vec{B_y}~~~~(2)\]\[\vec{A_y}=0\text{ since } \vec A\text{ has no vertical component}\]

Answer 4

other components for \(\vec A \text{ and }\vec B\) as follows...

Answer 5

\[|\vec{A_x}|=150\cos0^\circ=150(1)=150\]\[|\vec{B_x}|=300\cos60^\circ=300(0.5)=150\]\[|\vec{B_y}|=300\sin60^\circ=300(0.866)=259.81\]for equation (1):\[|\vec F|\cos\theta=150+150=300~~~~(1)\]for equation (2):\[|\vec F|\sin\theta=259.81~~~~(2)\](2) divided by (1):\[\frac{\bcancel{|\vec F|}\sin\theta}{\bcancel{|\vec F|}\cos\theta}=\frac{259.81}{300}\]\[\tan\theta=0.86602540378444\]\[\theta=\tan^{-1}(0.86602540378444)\]\[\theta=40.89^\circ\]And for the magnitude of \(\vec F\), either (1) or (2) can be used. Using (1):\[|\vec F|=\frac{300}{\cos40.89^\circ}=\frac{300}{0.756}\]\[|\vec F|=396.863\]and so,\[\vec F=396.863\angle40.89^\circ\]which is the same as b solution....