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goblinko [34]
3 years ago
11

Ball joints on a vehicle equipped with MacPherson struts are being inspected for wear. Which of the following would be the corre

ct method to check the load bearing joint?
Question 2 options:

Leave the vehicle weight on the tires, turn the steering wheel and listen for a popping noise.


Raise the vehicle by the subframe and measure the joints vertical movement.


Raise the vehicle by the lower control arm and measure the joints vertical movement.


Raise the vehicle by the subframe and turn the steering wheel and listen for a popping noise.
Engineering
1 answer:
Aleks04 [339]3 years ago
5 0

he following would be the correct method to check the load bearing joint is Raise the vehicle by the subframe and turn the steering wheel and listen for a popping noise.

Explanation:

  • Ball joints on a vehicle equipped with MacPherson struts are being inspected for wear. It can be checked by raising the vehicle by the subframe and turn the steering wheel and listen for a popping noise.
  • When a joint of CV becomes rusted or damaged, the flexibility of axle becomes out of order and starts making a constant clicking noise whenever the wheels are turned around.
  • The struts are the most important part of the suspension system.
  • The metallic clunking noise is found to be one of the most noticeable and frequent symptoms that shows of a bad ball joint which gives a bad clunking or knocking noise whenever the suspension operates in an up and down motion.
  • The ball joint can be broken in the following ways:
  • the ball detaching from the socket and stud breakage, nonetheless of the type of breakage, but it is catastrophic.
  • When the joint of the ball completely breaks down, the wheel becomes free to move in all the directions.

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Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following
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Given:

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The transition from the lamina to turbulent begins when the critical Reynolds

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(a).  \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16  \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\

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Answer:

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a=\sqrt{2^2+54^2}\ ft/s^2

a=54.43\ ft/s^2

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