A force vector f1 has a magnitude of 50 north it makes a 45 degree angle with a positive x axis. F2 has a magnitude of 70 north and points straight down. Determine the vector that results from adding f1 and f2
are "north" and "down" related to the x,y,z axes?
I think?
I’m not sure. I just started this course. I think so. Do you know how to solve it or get it started?
ok - I think "north" is a typo, maybe for Newtons
That means that
F1 = 50 cos45° i + 50sin45° j = 35.35i + 35.35j
F2 = -70j
So F1+F2 = 35.35i - 34.65j
Now, that's almost the same as 35.35i - 35.35j
which would be 50N at an angle of -45° to the x-axis.
To be more exact, it would be
50/√2 i + (50/√2 - 70)j or
49.50N at an angle of θ where
tanθ = (50/√2 - 70) / (50/√2) = -0.9799
θ = -44.418°
To determine the vector that results from adding f1 and f2, we need to find the components of each vector in the x and y directions, and then add them together.
Let's break down the given information:
- f1 has a magnitude of 50 N and makes a 45-degree angle with the positive x-axis.
- f2 has a magnitude of 70 N and points straight down.
To find the components of f1, we can use trigonometry. We can split f1 into its x and y components using the following formulas:
f1x = f1 * cos(θ)
f1y = f1 * sin(θ)
Where θ is the angle made by f1 with the positive x-axis.
For f1, we have:
f1x = 50 * cos(45°)
f1x = 50 * (√2 / 2) ≈ 35.36 N
f1y = 50 * sin(45°)
f1y = 50 * (√2 / 2) ≈ 35.36 N
Since f2 points straight down, its x-component is 0, and its y-component is -70.
Now, we can find the resulting vector by adding the components of f1 and f2:
Resultant vector, Rx = f1x + f2x
Resultant vector, Ry = f1y + f2y
Rx = 35.36 + 0
Rx = 35.36 N
Ry = 35.36 - 70
Ry = -34.64 N
The resulting vector (R) has a magnitude of 35.36 N in the x-direction and -34.64 N in the y-direction.