What equation do I use to find this? Thanks
The x and y components of a certain force are measured and found to be Fx = 68 ± 3N and Fy = 42 ± 2 N, respectively. Calculate the error associated with the magnitude of this force?
Well Magnitude is
sqrt((Fx^2 + Fy^2))
where:
Fx is the sum of all forces in x-direction
Fy is sum of all forces in y-direction
Solution:
Fx = 68 + 3 = 71N
Fy= 42 + 2 = 44 N
Magnitutde:
sqrt((71^2+44^2)) = 83.53
Fx=68 - 3 = 65N
Fy = 42 - 2 = 40 N
Magnitude:
sqrt((65^2+40^2)) = 76.32 N
For error:
(Max Force - Min Force)/(Max Force) * 100 = % error
(83.53 - 76.32)/(83.53) * 100 = 8.63%
Which means The magnitude of the addition part is 8.63% bigger than the subtraction part.
To find the error associated with the magnitude of the force, you can use the formula for error propagation. In this case, since the magnitude of the force (F) is given by the square root of the sum of the squares of the x and y components (Fx and Fy), the formula becomes:
ΔF = sqrt((ΔFx/F)^2 + (ΔFy/F)^2) * F
Where:
- ΔF is the error associated with the magnitude of the force
- ΔFx and ΔFy are the errors associated with the x and y components of the force, respectively
- Fx and Fy are the x and y components of the force
- F is the magnitude of the force
By substituting the given values into the formula, you can calculate the error associated with the magnitude of the force as follows:
ΔF = sqrt((3/68)^2 + (2/42)^2) * √(68^2 + 42^2)
Simplify the expression inside the square root:
ΔF = sqrt((0.0441 + 0.0119) * 68^2 + 42^2)
ΔF = sqrt(0.056) * 68^2 + 42^2
ΔF = 0.236 * 68^2 + 1764
Finally, you can calculate the error associated with the magnitude of the force.