Answer:
The value of bending stress on the pinion 35.38 M pa
Explanation:
Given data
m = 2 mm
Pressure angle
= 20°
No. of teeth T = 17
Face width (b) = 20 mm
Speed N = 1650 rpm
Power = 1200 W
Diameter of the pinion gear
D = m T
D = 2 × 17
D = 34 mm
Velocity of the pinion gear
![V =\pi D( \frac{N}{60} )](https://tex.z-dn.net/?f=V%20%3D%5Cpi%20D%28%20%5Cfrac%7BN%7D%7B60%7D%20%29)
![V = 3.14 (0.034) \frac{(1650)}{60}](https://tex.z-dn.net/?f=V%20%3D%203.14%20%280.034%29%20%5Cfrac%7B%281650%29%7D%7B60%7D)
![V = 2.93 \frac{m}{s}](https://tex.z-dn.net/?f=V%20%3D%202.93%20%5Cfrac%7Bm%7D%7Bs%7D)
Form factor for the pinion gear is
Y = 0.303
Now
![K_{v} = \frac{6.1 +0.303}{6.1} = 1.049](https://tex.z-dn.net/?f=K_%7Bv%7D%20%3D%20%5Cfrac%7B6.1%20%2B0.303%7D%7B6.1%7D%20%3D%201.049)
Force on gear tooth
![F = \frac{P}{V}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BP%7D%7BV%7D)
![F = \frac{1200}{2.93}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B1200%7D%7B2.93%7D)
F = 408.73 N
Now the bending stress is given by the formula
![\sigma = \frac{K_{v} F}{m b y}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cfrac%7BK_%7Bv%7D%20F%7D%7Bm%20b%20y%7D)
![\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cfrac%7B%281.049%29%28408.73%29%7D%7B%280.002%29%280.02%29%280.303%29%7D)
= 35.38 M pa
This is the value of bending stress on the pinion
I think its Mercury because it's the closest to the sun.
Let us consider two bodies having masses m and m' respectively.
Let they are separated by a distance of r from each other.
As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -
where G is the gravitational force constant.
From the above we see that F ∝ mm' and ![F\alpha \frac{1}{r^{2} }](https://tex.z-dn.net/?f=F%5Calpha%20%5Cfrac%7B1%7D%7Br%5E%7B2%7D%20%7D)
Let the orbital radius of planet A is
= r and mass of planet is
.
Let the mass of central star is m .
Hence the gravitational force for planet A is ![f_{1} =G \frac{m_{1}*m }{r^{2} }](https://tex.z-dn.net/?f=f_%7B1%7D%20%3DG%20%5Cfrac%7Bm_%7B1%7D%2Am%20%7D%7Br%5E%7B2%7D%20%7D)
For planet B the orbital radius
and mass
Hence the gravitational force ![f_{2} =G\frac{m m_{2} }{r^{2} }](https://tex.z-dn.net/?f=f_%7B2%7D%20%3DG%5Cfrac%7Bm%20m_%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D)
![f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }](https://tex.z-dn.net/?f=f_%7B2%7D%20%3DG%5Cfrac%7Bm%2A3m_%7B1%7D%20%7D%7B%5B2r_%7B1%7D%5D%20%5E%7B2%7D%20%7D)
![= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B3%7D%7B4%7D%20G%5Cfrac%7Bmm_%7B1%7D%20%7D%7Br_%7B1%7D%20%5E%7B2%7D%20%7D)
Hence the ratio is ![\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }](https://tex.z-dn.net/?f=%5Cfrac%7Bf_%7B2%7D%20%7D%7Bf_%7B1%7D%20%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B3%7D%7B4%7DG%20mm_%7B1%2Fr_%7B1%7D%20%5E2%7D%20%20%7D%7BGmm_%7B1%7D%2Fr_%7B1%7D%20%5E2%20%7D)
[ ans]
So he has 7 sheepleft if i did it correctly
15-8=7