Question #7d3d7

1 Answer
Morgan
Dec 16, 2017

See below.

Explanation:

We can break the given vectors up into their components using trigonometry.

Beginning with vector A, we know #abs(vecA)=8.00"m"#.

In other words, the magnitude, or "length," of the vector is #8.00"m"#. We see that #vecA# is oriented in such a way that it has no horizontal component, i.e. it makes an angle of #-90^o# with respect to the positive x-axis. Therefore, without having to use any trigonometry, we know that #color(blue)(A_y=-8.00"m")# and #color(blue)(A_x=0.00"m")#.

We are given that #abs(vecB)=15.0"m"# and #theta_2=31^o#, measured from the positive y-axis. This vector has both horizontal (parallel) and vertical (perpendicular) components. From the diagram below, we see that:

#sin(theta)="opposite"/"hypotenuse"#

#=>sin(theta_2)=B_x/B#

Solve for #B_x#:

#B_x=Bsin(theta_2)#

#=>=(15.0)sin(31^o)#

#=>color(blue)(B_x~~7.73"m")#

Similarly, we can use the cosine function to find #B_y#.

#B_y=(15.0)cos(31^o)#

#=>color(blue)(B_y~~12.9"m")#

From OP

The other components are found in the same way.

#D_x=Dsin(theta_4)#

#=>=(-10.0)sin(51^o)#

#=>color(blue)(D_x~~-7.77"m")#

#D_y=(10.0)cos(51^o)#

#=>color(blue)(D_y~~6.29"m")#

#C_x=Ccos(theta_3)#

#=>=(-12.0)cos(23^o)#

#=>color(blue)(C_x~~-11.0"m")#

#C_y=(-12.0)sin(23^o)#

#=>color(blue)(C_y~~-4.69"m")#

Note that the negative signs indicate direction, and follow from the position of the vectors with respect to the chosen coordinate system, i.e. the x-y plane, with conventional directions chosen (down and left negative, up and right positive).

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