Using a force table the direction and magnitude had to be found for the following measurements:
F1 = 200 grams at 30 degrees; F2 = 200 grams at 120 degrees
The magnitude and direction found was 280 grams at 257 degrees.
Using the magnitude and direction of F1 and F2 I need to find out the magnitude and direction of the resultant vector F1-F2.
I am not sure how to do this. Thank you for you help.
the force that was needed to balance out F1 and F2 is equal in magnitude to the resultant, but opposite in direction
so the resultant would be 280g at 77º
since F1 and F2 have equal magnitudes and are 90º apart; the expected resultant would be 283g at 75º __ so your outcome looks good
Oh ok I got 283 and 75 degreed from usng the Rx an Ry equations and then doing tan^-1(Ry/Rx), so 283 and 75 is my answer?
To find the magnitude and direction of the resultant vector F1-F2, we can use the concepts of vector addition.
1. Start by finding the horizontal and vertical components of F1 and F2.
For F1:
- Horizontal Component (Fx1): F1 * cosθ1
Fx1 = 200 grams * cos(30°)
- Vertical Component (Fy1): F1 * sinθ1
Fy1 = 200 grams * sin(30°)
For F2:
- Horizontal Component (Fx2): F2 * cosθ2
Fx2 = 200 grams * cos(120°)
- Vertical Component (Fy2): F2 * sinθ2
Fy2 = 200 grams * sin(120°)
2. Add the horizontal and vertical components of F1 and F2 separately to get the components of the resultant vector.
Resultant Horizontal Component (FRx): Fx1 - Fx2
Resultant Vertical Component (FRy): Fy1 - Fy2
3. Use the resultant components to find the magnitude and direction of the resultant vector using the Pythagorean theorem and inverse tangent function:
Resultant Magnitude (FR): sqrt(FRx^2 + FRy^2)
Resultant Direction (θR): atan(FRy / FRx)
Now, let's calculate the magnitude and direction of the resultant vector based on the given values for F1 and F2.
Given:
F1 = 200 grams at 30 degrees
F2 = 200 grams at 120 degrees
1. Calculate the horizontal and vertical components of F1 and F2:
For F1:
Fx1 = 200 grams * cos(30°)
= 200 grams * 0.866
≈ 173.2 grams (rounded to one decimal place)
Fy1 = 200 grams * sin(30°)
= 200 grams * 0.5
= 100 grams
Similarly, for F2:
Fx2 = 200 grams * cos(120°)
= -200 grams * 0.5
= -100 grams
Fy2 = 200 grams * sin(120°)
= 200 grams * 0.866
≈ 173.2 grams (rounded to one decimal place)
2. Add the components of F1 and F2:
FRx = Fx1 - Fx2
= 173.2 grams - (-100 grams)
= 273.2 grams
FRy = Fy1 - Fy2
= 100 grams - 173.2 grams
= -73.2 grams
3. Calculate the magnitude and direction of the resultant vector:
FR = sqrt(FRx^2 + FRy^2)
= sqrt((273.2 grams)^2 + (-73.2 grams)^2)
≈ sqrt(74822.24 grams^2)
≈ 273.3 grams (rounded to one decimal place)
θR = atan(FRy / FRx)
= atan(-73.2 grams / 273.2 grams)
≈ atan(-0.2681)
≈ -14.9 degrees (rounded to one decimal place)
Therefore, the magnitude of the resultant vector F1-F2 is approximately 273.3 grams, and the direction is approximately -14.9 degrees.