If /U/=9, /V/=11, and theta = 43 deg (angle between), find the magnitude of U+V and give the direction that the that the resultant makes with U.
make a sketch by completing the usual parallelogram , if U+V = R
R^2 = 11^2 + 9^2 - 2(9)(11)cos 137°
= 346.808
R = 18.62
now use the Sine Law to findØ, the angle between V and R.
sinØ/9 = sin137°/18.62
sinØ = .329595
Ø = 19.24° with vector V, then
23.76° with vector U
To find the magnitude of U+V, we can use the Law of Cosines. According to the Law of Cosines, the magnitude of U+V can be calculated as:
|U+V|^2 = |U|^2 + |V|^2 - 2|U||V|cos(theta)
Given that |U| = 9, |V| = 11, and theta = 43 degrees, we can substitute these values into the equation:
|U+V|^2 = 9^2 + 11^2 - 2(9)(11)cos(43)
Now, we can calculate the magnitude of U+V:
|U+V|^2 = 81 + 121 - 198cos(43)
|U+V|^2 = 202 - 198cos(43)
Since we are asked to find the magnitude, we can take the square root of both sides:
|U+V| = √(202 - 198cos(43))
Using a calculator, we can find:
|U+V| ≈ 11.29
Therefore, the magnitude of U+V is approximately 11.29.
Next, to find the direction that the resultant makes with U, we can use the Law of Sines. According to the Law of Sines, the direction (theta_r) can be calculated as:
sin(theta_r)/|V| = sin(theta)/|U+V|
We know that sin(theta) = sin(43) and |U+V| ≈ 11.29. Substituting these values, we have:
sin(theta_r) / 11 = sin(43) / 11.29
Next, we can solve for sin(theta_r) by cross multiplying:
sin(theta_r) = (sin(43) / 11.29) * 11
sin(theta_r) ≈ 0.690
To find the value of theta_r, we can take the inverse sin of both sides:
theta_r ≈ arcsin(0.690)
Using a calculator, we find:
theta_r ≈ 43 degrees
Therefore, the direction that the resultant makes with U is 43 degrees.
To solve this problem, we can use vector addition and trigonometry.
1. Find the x and y components of vector U:
- U has a magnitude of 9, so the x-component (Ux) is 9 * cos(0°) = 9 * 1 = 9.
- The y-component (Uy) is 9 * sin(0°) = 9 * 0 = 0.
2. Find the x and y components of vector V:
- V has a magnitude of 11, so the x-component (Vx) is 11 * cos(43°).
- The y-component (Vy) is 11 * sin(43°).
Using trigonometric functions, we can find the values of Vx and Vy:
Vx = 11 * cos(43°)
= 11 * 0.7314
≈ 8.045
Vy = 11 * sin(43°)
= 11 * 0.6819
≈ 7.501
3. Add the corresponding components of U and V to find the resultant vector R:
Rx = Ux + Vx = 9 + 8.045 ≈ 17.045
Ry = Uy + Vy = 0 + 7.501 ≈ 7.501
4. Calculate the magnitude of the resultant vector R using the Pythagorean theorem:
|R| = √(Rx^2 + Ry^2) ≈ √(17.045^2 + 7.501^2) ≈ √(289.898 + 56.26) ≈ √346.158 ≈ 18.61
Therefore, the magnitude of U + V is approximately 18.61.
5. To find the direction of the resultant vector R relative to vector U, we can use inverse trigonometric functions:
Let's find the angle θ between vector U and R:
θ = arctan(Ry / Rx) ≈ arctan(7.501 / 17.045) ≈ arctan(0.4396)
θ ≈ 23.86°
So, the resultant vector R makes an angle of approximately 23.86 degrees with vector U.