Two vectors F1 and F2 with magnitude F1=42N and F2=70N and where angle1=83 and angle2= 300. The angles are measures from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vector where F=F1+F2?
F = 42N[83o] + 70N[300o].
X = 42*Cos83 + 70*Cos300 =
Y = 42*sin83 + 70*sin300 =
Tan A = Y/X, A = ?
F = X/Cos A, or F = Y/sin A.
To find the magnitude of a resultant vector, we can use vector addition. In this case, we need to add vectors F1 and F2 to find the resultant vector F.
To add two vectors, we need to resolve them into their horizontal and vertical components. Let's start by finding the horizontal and vertical components of F1:
F1x = F1 * cos(angle1)
F1y = F1 * sin(angle1)
Now let's find the horizontal and vertical components of F2:
F2x = F2 * cos(angle2)
F2y = F2 * sin(angle2)
Next, we add the horizontal and vertical components of F1 and F2 to find the horizontal and vertical components of the resultant vector F:
Fx = F1x + F2x
Fy = F1y + F2y
Finally, we can calculate the magnitude of the resultant vector F using the Pythagorean theorem:
F = √(Fx^2 + Fy^2)
Plugging in the values:
F1 = 42 N
F2 = 70 N
angle1 = 83 degrees
angle2 = 300 degrees
F1x = 42 * cos(83)
F1y = 42 * sin(83)
F2x = 70 * cos(300)
F2y = 70 * sin(300)
Fx = F1x + F2x
Fy = F1y + F2y
F = √(Fx^2 + Fy^2)
Now you can calculate the values and find the magnitude of the resultant vector F.