Answer:
w = 11.211 KN/m
Explanation:
Given:
diameter, d = 50 mm
F.S = 2
L = 3
Due to symmetry, we have:
![Ay = By = \frac{w * 6}{2} = 3w](https://tex.z-dn.net/?f=%20Ay%20%3D%20By%20%3D%20%5Cfrac%7Bw%20%2A%206%7D%7B2%7D%20%3D%203w%20)
![P_c_r = 3w * F.S = 3w * 2.0 = 6w](https://tex.z-dn.net/?f=%20P_c_r%20%3D%203w%20%2A%20F.S%20%3D%203w%20%2A%202.0%20%3D%206w%20)
![I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7](https://tex.z-dn.net/?f=%20I%20%3D%20%5Cfrac%7B%5Cpi%7D%7B64%7D%20%280.05%29%5E4%20%3D%203.067%2A10%5E-%5E7%20)
To find the maximum intensity, w, let's take the Pcr formula, we have:
![P_c_r = \frac{\pi^2 E I}{(KL)^2}](https://tex.z-dn.net/?f=%20P_c_r%20%3D%20%5Cfrac%7B%5Cpi%5E2%20E%20I%7D%7B%28KL%29%5E2%7D%20)
Let's take k = 1
Substituting figures, we have:
![6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}](https://tex.z-dn.net/?f=%206w%20%3D%20%5Cfrac%7B%5Cpi%5E2%20%2A%20200%2A10%5E9%20%2A%203.067%2A10%5E-%5E7%7D%7B%281%20%2A%203%29%5E2%7D%20)
Solving for w, we have:
![w = \frac{67266.84}{6}](https://tex.z-dn.net/?f=%20w%20%3D%20%5Cfrac%7B67266.84%7D%7B6%7D%20)
w = 11211.14 N/m = 11.211 KN/m
Since Area, A= pi * (0.05)²
. This means it is safe
The maximum intensity w = 11.211KN/m
Answer:
Part a)
Moment of inertia of the cylinder is given as
![I = 1.21 \times 10^{-4} kg m^2](https://tex.z-dn.net/?f=I%20%3D%201.21%20%5Ctimes%2010%5E%7B-4%7D%20kg%20m%5E2)
Part B)
Height of the cylinder is of no use here to calculate the inertia
Part C)
Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as ![I = 1/2 mR^2](https://tex.z-dn.net/?f=I%20%3D%201%2F2%20mR%5E2)
Explanation:
As we know that the inclined plane is of length L = 3 m
and its inclination is given as 25 degree
so we know that acceleration of center of mass of the cylinder is constant so we will have
![v_f^2 = v_i^2 + 2 a L](https://tex.z-dn.net/?f=v_f%5E2%20%3D%20v_i%5E2%20%2B%202%20a%20L)
so we have
![v_f^2 = 0 + 2a(3)](https://tex.z-dn.net/?f=v_f%5E2%20%3D%200%20%2B%202a%283%29)
now we know that
![v_{avg} = \frac{L}{t} = \frac{v_f + v_i}{2}](https://tex.z-dn.net/?f=v_%7Bavg%7D%20%3D%20%5Cfrac%7BL%7D%7Bt%7D%20%3D%20%5Cfrac%7Bv_f%20%2B%20v_i%7D%7B2%7D)
![\frac{3}{1.50} = \frac{v_f + 0}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B1.50%7D%20%3D%20%5Cfrac%7Bv_f%20%2B%200%7D%7B2%7D)
![v_f = 4 m/s](https://tex.z-dn.net/?f=v_f%20%3D%204%20m%2Fs)
Now we have know that final speed of the cylinder due to pure rolling is given as
![v_f = \sqrt{\frac{2gH}{1 + \frac{I}{mR^2}}}](https://tex.z-dn.net/?f=v_f%20%3D%20%5Csqrt%7B%5Cfrac%7B2gH%7D%7B1%20%2B%20%5Cfrac%7BI%7D%7BmR%5E2%7D%7D%7D)
![4 = \sqrt{\frac{2(9.81)(3 sin25)}{1 + \frac{I}{0.215(0.0319)^2}}}](https://tex.z-dn.net/?f=4%20%3D%20%5Csqrt%7B%5Cfrac%7B2%289.81%29%283%20sin25%29%7D%7B1%20%2B%20%5Cfrac%7BI%7D%7B0.215%280.0319%29%5E2%7D%7D%7D)
![I = 1.21 \times 10^[-4} kg m^2](https://tex.z-dn.net/?f=I%20%3D%201.21%20%5Ctimes%2010%5E%5B-4%7D%20kg%20m%5E2)
Part B)
Height of the cylinder is of no use here to calculate the inertia
Part C)
Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as ![I = 1/2 mR^2](https://tex.z-dn.net/?f=I%20%3D%201%2F2%20mR%5E2)
It would be D (hoping the picture you posted is what you needed help with
"Gauss's Law. The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field. ," Source: <span>hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html
If you would like more info please look at the website. Im only in middle school, so I am sorry if this is not what you were looking for.....</span>