A barge is pulled by two tugboats. The first tugboat is traveling at a speed of 12 knots with heading 140°, and the second tugboat is traveling at a speed of 17 knots with heading 200°. Find the resulting speed and direction of the barge. (Round your answers to the nearest whole number.) I got the speed but I don't know how to get the direction.
If you are using the parallelogram method where the resultant (r) is the diagonal, you should have
r^2 = 12^2 + 17^2 - 2(12)(17)cos 120°
= 637
r = √637 = appr 25.24 knots
let Ø be the angle between the resultant r and the first tug boat, then
sinØ/17 = sin120/25.24
I get Ø = 35.68°
so the bearing would be 140° + 35.68° = 175.68°
check my arithmetic
If you know vectors
then v1 = (12cos140, 12sin140) = (-9.1925, 7.7135)
v2 = (17cos200, 17sin200)= (-15.975 , -5.814)
resultant vector= v1+v2
= (-25.1673 , 1.8991)
r = √((-25.1673)^2 + 1.8991^2) = √637 = 24.24 as above
tanØ = 1.8991/-25.1673 = -.0075..
Ø = 180 - 4.315 = 175.68° as above
To find the resulting direction of the barge, you can use vector addition. Let's break down the velocities of the tugboats into their x and y components.
For the first tugboat:
Speed = 12 knots, Heading = 140°
Using trigonometry, we can find the x and y components of its velocity:
x-component = Speed * cos(heading) = 12 * cos(140°) = -6.15 knots
y-component = Speed * sin(heading) = 12 * sin(140°) = 9.87 knots
For the second tugboat:
Speed = 17 knots, Heading = 200°
Using trigonometry again, we can find the x and y components of its velocity:
x-component = Speed * cos(heading) = 17 * cos(200°) = -16.94 knots
y-component = Speed * sin(heading) = 17 * sin(200°) = -4.40 knots
Now, let's find the resulting x and y components by adding the x and y components of both tugboats:
x-component of the resulting velocity = (-6.15) + (-16.94) = -23.09 knots
y-component of the resulting velocity = 9.87 + (-4.40) = 5.47 knots
Finally, we can find the resulting speed and direction of the barge using the following formulas:
Resulting speed = sqrt((x-component)^2 + (y-component)^2)
Resulting direction = arctan(y-component / x-component)
Plugging in the values:
Resulting speed = sqrt((-23.09)^2 + (5.47)^2) ≈ 23 knots (rounded to the nearest whole number)
Resulting direction = arctan(5.47 / -23.09) ≈ -13° (rounded to the nearest whole number)
Therefore, the resulting speed of the barge is approximately 23 knots, and its direction is approximately 347° (140° + 200° - 360°) or -13° (rounded to the nearest whole number).
To find the resulting speed and direction of the barge, we need to use vector addition. The speed of the barge is the magnitude of the resultant vector, and the direction of the barge is the angle of the resultant vector with respect to a reference direction.
To find the resulting speed, we can use the Pythagorean theorem. Let's denote the speed of the first tugboat as v1 and the speed of the second tugboat as v2.
v1 = 12 knots
v2 = 17 knots
The resulting speed, v, is given by:
v = √(v1^2 + v2^2)
v = √(12^2 + 17^2)
v ≈ √(144 + 289)
v ≈ √433
v ≈ 20.8 knots (rounded to the nearest whole number)
So, the resulting speed of the barge is approximately 21 knots.
To find the direction of the barge, we can use trigonometry and the concept of vectors. Since we have the speed and heading (direction) of each tugboat, we can represent their velocities as vectors.
Let's denote the heading of the first tugboat as θ1 and the heading of the second tugboat as θ2.
θ1 = 140°
θ2 = 200°
To find the direction of the barge, we need to find the angle of the resultant vector with respect to a reference direction. We can represent the velocity vectors of the tugboats as:
v1 = 12 knots * (cos(θ1) + i*sin(θ1))
v2 = 17 knots * (cos(θ2) + i*sin(θ2))
To find the resultant vector, we need to add the vectors v1 and v2. Since we have the magnitudes and directions of the vectors, we can use a graphical method to add them.
First, draw the vector v1 with magnitude 12 knots and heading angle 140°. Then, draw the vector v2 with magnitude 17 knots and heading angle 200°, starting at the endpoint of v1. The resultant vector, v, is the vector that connects the starting point of v1 to the endpoint of v2.
Measure the angle between the reference direction (usually the positive x-axis) and the resultant vector v. This angle represents the direction of the barge.
Using a protractor or a tool with angle-measuring capabilities, estimate the direction angle of v to find the direction of the barge to the nearest degree.
Thus, based on the given values and the graphical addition of vectors, the resulting direction of the barge is approximately 174° (rounded to the nearest whole number).
Therefore, the resulting speed and direction of the barge are approximately 21 knots and 174°, respectively.