Answer:
2 N / m
Explanation:
It is necessary to determine the maximum w so that the normal stress in AB and CD rods does not exceed the allowable stress.
The cross-section of respective rods are:
A_ab = 10 mm^2
A_cd = 15 mm^2
Step 1 :The maximum load is determined from the condition that the normal stress (б)
is not higher than (б_allow).
б_ab = F_ab / A_ab =< б_allow
б_cd = F_cd / A_cd =< б_allow
Step 2 : Compute the static size
We apply the Equilibrium conditions:
1) Sum of Moments (A) = 0
-w * 6 * 0.5 * 6 * 0.75 + 6*F_cd = 0
F_cd = (13.5 / 6) * w = 2.25*w KN
2) Sum of Forces (y-direction) = 0
F_cd + F_ab - 0.5*6*w = 0
F_ab = 3*w - 2.25*w = 0.75*w KN
Step 3 : Compute Load w
1) Sector AB :
F_ab / A_ab =< б_allow
0.75*w / 10 * 10^-6 =< 300KPa
0.75*w =< 3
w =< 4 N / m
2) Sector CD :
F_cd / A_cd =< б_allow
2.25*w / 15 * 10^-6 =< 300KPa
2.25*w =< 4.5
w =< 2 N / m
Answer: w = 2 N / m
Answer:
105 m/s
Explanation:
Given that the speed of train A, = 45 m/s from west to east.
Speed of train B, = 60 m/s from east to west.
Train B is moving in the opposite direction with respect to the speed of train A. Assuming that the speed from east to west direction is positive.
So, the speed of train A from east to west= - 45 m/s
The speed of train B w.r.t train A m/s
Hence, the speed of train B w.r.t train A is 105 m/s from east to west.
Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a).
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now,
A' = amplitude = 1.4142 m
b).
m' = 2m
Hence,
c).
Therefore, factor
Thus, the energy will change half times as the result of the collision.