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HW#2b part 2 Page 1 of 3 8. - SFSU Physics & Astronomywman/phy111hw/HW solutions/hw2b_part2.pdf ·...
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HW#2b part 2 Page 1 of 3 8. 9. 10. Practice with vectors: (important!) For each of the vectors below, I.) do a rough sketch of it II.) convert it to the form it is not already in (size&direction, or components) (Note: cw=clockwise, ccw=counterclockwise) a.) = 5 m, 30 o ccw from +x axis D x = m D y = m y x A r A y A x θ
Transcript of HW#2b part 2 Page 1 of 3 8. - SFSU Physics & Astronomywman/phy111hw/HW solutions/hw2b_part2.pdf ·...
HW#2b part 2 Page 1 of 3
8.
9.
10. Practice with vectors: (important!) For each of the vectors below, I.) do a rough sketch of it II.) convert it to the form it is not already in (size&direction, or components) (Note: cw=clockwise, ccw=counterclockwise)
a.) = 5 m, 30o ccw from +x axis
Dx = m
Dy = m
y
x
Ar
Ay
Ax
θ
HW#2b part 2 Page 2 of 3
Answers. Here and later we will use the relations between the (magnitude, angle) and component representations of a vector:
2 2
cossintan
x yx
yy
x
A A A A AA A AA
θθθ
⎫= +=⎪ ⎧
⎬ ⎨ == ⎩⎪⎭
So for part (a),
( )( )
cos 5 m cos30 4.33 m
D sin 5 m sin 30 2.5 mx
y
D D
D
θ
θ
= = ° =
= = ° =
b.) = 15 m/s, 10o cw from –y axis vav x = -2.604 m/s vav y = -14.77 m/s Answer. Look at the diagram - the angle, measured in the usual direction from the positive x axis, is equal to 260 °. And both x and y components of the vector are negative. Let's see if the trig functions get this right automatically.
( )( )
cos 15 m/s cos 260 2.604 m/s
sin 15 m/s sin 260 14.77 m/sx
y
A A
A A
θ
θ
= = ° = −
= = ° = −
c.) Qx = -6 m, Qy = +2 m
= 6.325m, 18.43 o cw from –x axis Answer. Here we make the transformation back from components to magnitude and angle.
( ) ( )
2 2
2 26 m 2 m
6.325 m
x yQ Q Q= +
= − +
=
( )( )
2 mtan .33333
6 m18.43 180 161.57
y
x
θ
θ
= = = −−
= − + = °
The angle called for is not the usual angle θ, but is equal to 180° minus the standard angle. So the answer is 180°-161.57° = 18.43°.
y
x
5 m
30 ° Ay
Ax
y
x
15 m/s 10 °
Ay
Ax
θ=260 °
180°-θ
y
x
θ Qy 2 m
Qx -6 m
HW#2b part 2 Page 3 of 3
11. Use unit vector notation to express each of the vectors in Figure 3-39, in which the magnitudes of A, B, C, and D are respectively given by 36 m, 48 m, 24 m, and 36 m. A =( 27.6 x + 23.1 y) m B =( 45.4 x + (-15.6) y) m C =( -21.8 x + 10.1 y) m D =( 36 x + 0 y) m
Notice that here A, B, C and D and also x and y are bold, meaning they are vectors. Ax= A*cos (40) = 27.6 Ay=A*sin(40) = 23.1 Bx= B*cos (19) = 45.4 By= - B*sin(19) = -15.6 Cx= - C*cos (25) = - 21.8 Cy= C *sin(25) = 10.1 Dx= D*cos (90) = 0 Dy= D*sin(90) = D = 36 THESE ARE EXTREMELY IMPORTANT. PLEASE PRACTICE AGAIN AND AGAIN.
