Explore BrainMass

# Vector

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

1) A team of surveyors mark off distances represented by vectors of (85m, 55 degrees), (43m, 0 degrees) and (37m, 345 degrees). Solve for the resultant displacement.

2) Find the resultant force. (463 N, 137 degrees), (258 N, 268 degrees) and (379 N, 128 degrees)

https://brainmass.com/math/basic-algebra/solve-resultant-displacement-31041

## SOLUTION This solution is FREE courtesy of BrainMass!

A team of surveyors mark off distances represented by vectors of (85m, 55 degrees), (43m,0 degrees)and (37m,345 degrees). Solve for the resultant displacement.

Component is the horizontal (x) direction = (Magnitude) x (Cos phi )
Component is the vertical (y) direction = (Magnitude) x (Sin phi )

a) (85m, 55 degrees)
Magnitude = 85 m
phi = 55 degrees
x component = 48.75 m =85 Cos55
y component = 69.63 m =85 Sin55

b) (43m,0 degrees)
Magnitude = 43 m
phi = 0 degrees
x component = 43 m =43 Cos0
y component = 0 m =43 Sin0

c) (37m, 345 degrees)
Magnitude = 37 m
phi = 345 degrees
x component = 35.74 m =37 Cos345
y component = -9.58 m =37 Sin345

Total displacement in x direction= 127.49 m =48.75+43+35.74
Total displacement in y direction= 60.05 m =69.63+0+-9.58

Resultant displacement = square root of {(x component) ^2 + (y component)^2}= 140.92 m =square root of (127.49^2+60.05^2)

Angle between the resultant and x axis = tan inverse ( y component / x component)

25.22 degrees = tan inverse (60.05/127.49)

Resultant displacement = 140.92 m at an angle of 25.22 degrees from x axis
0r (140.92m, 25.22degrees)

Qn 2) Find the resultant forces. (463 N,137 degrees),(258 N,268 degrees) and (379 N 128 degrees)

Component is the horizontal (x) direction = (Magnitude) x (Cos phi )
Component is the vertical (y) direction = (Magnitude) x (Sin phi )

a) (463N, 137 DEGREES)
Magnitude = 463 N
phi = 137 degrees
x component = -338.62 N =463 Cos137
y component = 315.77 N =463 Sin137

b) (258 N, 268 DEGREES)
Magnitude = 258 N
phi = 268 degrees
x component = -9.00 N =258 Cos268
y component = -257.84 N =258 Sin268

c) (379 N 128 DEGREES)
Magnitude = 379 N
phi = 128 degrees
x component = -233.34 N =379 Cos128
y component = 298.66 N =379 Sin128

Total force in x direction= -580.96 N =-338.62+-9+-233.34
Total force in y direction= 356.59 N =315.77+-257.84+298.66

Resultant force = square root of {(x component)^2 + (y component)^2}= 681.67 m =square root of (-580.96^2+356.59^2)

Angle between the resultant and x axis = tan inverse ( y component / x component)

-31.54 degrees = tan inverse (356.59/-580.96)
or 148.46 degrees =180-31.54