GIVEN:
Amplitude, A = 0.1mm
Force, F =1 N
mass of motor, m = 120 kg
operating speed, N = 720 rpm
= 
Formula Used:
Solution:
Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K
We know that:
so,
= 75.39 rad/s
Using the given formula:
Damping is negligible, so, 
will give the tranfer function
Therefore,
= 
=
Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm
Answer:
Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.
Explanation:
Solution
Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating pressure shortens the hose life.
In systems where pressure peaks are severe, select or pick a hose with higher maximum operating pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.
Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.
Answer:
a) 2∪p/lb (l+b)dH
b) po exp( 4∪x/l)
Explanation:
please check the attachment for proper explanation and proper sign notations thanks.
Answer:
as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.
Explanation:
4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum
