the total aerodynamic force F acting on the airplane has a magnitude of 6250 lb. Resolve this force into vertical and horizontal
components
1 answer:
To solve this problem, we need the figure to help us understand. I believe I found the correct figure for this problem (see attached pic).
We see that the resultant force acts at an angle of 6 degrees from the vertical axis. The resultant force is also equivalent to the hypotenuse of the triangle. Now to solve this problem, we need the trigonometric functions specifically the cos and sin functions. We know that:
sin θ = side opposite to the angle / hypotenuse = Fx / F
cos θ = side adjacent to the angle / hypotenuse = Fy / F
Therefore calculating for Fx and Fy:
Fx = F sin θ
Fx = 6250 lb (sin 6)
Fx = 653.3 lb
Fy = F cos θ
Fy = 6250 lb (cos 6)
Fy = 6215.76 lb
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Answer:
(a)
(b)
Explanation:
(a) The electric field for a point charge is given by the following formula:
Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.
Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.
In cylindrical coordinates, da = r dr dθ. So,
Hence, dE is now:
The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.
It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.
Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,
Therefore, vertical component of dE now becomes
Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:
(b) We will have a similar approach, but a simpler integral.
Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.