Determine the magnitude of the vector sum V=V1+V2 and the angle theta(x) which V makes counterclockwise from the positive x-axis.
V1 = 34 units and makes a 20 degree angle with the +x axis
V2 = 28 units and forms a 3,4,5 triangle with the -x axis
To determine the magnitude of the vector sum V=V1+V2, we can use the Pythagorean theorem. The magnitude of the vector sum is given by:
|V| = sqrt((V1)^2 + (V2)^2 + 2(V1)(V2)cos(theta))
where theta is the angle between V1 and V2.
Let's first determine theta.
V1 makes a 20 degree angle with the +x axis.
V2 forms a 3,4,5 triangle with the -x axis. Since the triangle is a right triangle, the angle between V2 and the -x axis is 90 degrees.
So, the angle between V1 and V2 is:
theta = 20 degrees + 90 degrees = 110 degrees
Now, substitute the values into the equation to calculate the magnitude of V:
|V| = sqrt((34)^2 + (28)^2 + 2(34)(28)cos(110 degrees))
Simplifying the equation gives:
|V| = sqrt(1156 + 784 + 1904cos(110 degrees))
Now, calculate the cosine of 110 degrees:
cos(110 degrees) = -0.342
Substitute this value into the equation:
|V| = sqrt(1156 + 784 + 1904*(-0.342))
Calculating the expression inside the square root gives:
|V| = sqrt(1156 + 784 - 650.368)
Simplifying further gives:
|V| = sqrt(2299.632)
Finally, calculating the square root gives:
|V| ≈ 47.97 units
Therefore, the magnitude of the vector sum V is approximately 47.97 units when rounded to two decimal places.
To determine the magnitude of the vector sum V = V1 + V2, we can use the Pythagorean theorem. The magnitude of V can be calculated using the formula:
|V| = sqrt(Vx^2 + Vy^2)
Where Vx and Vy are the x and y components of V respectively.
First, let's calculate the x and y components of V1:
V1x = V1 * cos(theta1)
V1y = V1 * sin(theta1)
Where theta1 is the angle V1 makes counterclockwise from the positive x-axis, which is given as 20 degrees.
V1x = 34 * cos(20°)
V1y = 34 * sin(20°)
Next, let's calculate the x and y components of V2:
V2x = V2 * cos(theta2)
V2y = V2 * sin(theta2)
Since V2 forms a 3-4-5 triangle with the -x axis, we can find the angle theta2 using trigonometry. The angle opposite the side with length 4 is 90 degrees, and the angle opposite the side with length 3 is theta2.
sin(theta2) = 3/5
theta2 = arcsin(3/5) ≈ 36.87°
V2x = 28 * cos(180° - 36.87°) (since it forms a triangle with -x axis)
V2y = 28 * sin(180° - 36.87°)
Now we can calculate the magnitude of V by adding the x and y components of V:
Vx = V1x + V2x
Vy = V1y + V2y
Finally, we can substitute these values into the magnitude formula:
|V| = sqrt(Vx^2 + Vy^2)
Using these steps, we can solve for the magnitude of the vector sum V and the angle theta(x) it makes counterclockwise from the positive x-axis.
V = 34[20o] + 28[53.1o]
X = 34*cos20 + 28*cos53.1 = 48.75 Units.
Y = 34*sin20 + 28*sin53.1 = 34.03 Units.
V^2=X^2 + Y^2 = 48.75^2 + 34.03^2=3535.
V = 59.5 Units.