Answer: The net force in every bolt is 44.9 kip
Explanation:
Given that;
External load applied = 245 kip
number of bolts n = 10
External Load shared by each bolt (P_E) = 245/10 = 24.5 kip
spring constant of the bolt Kb = 0.4 Mlb/in
spring constant of members Kc = 1.6 Mlb/in
combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6) = 0.4 / 2 = 0.2 Mlb/in
Initial pre load Pi = 40 kip
now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them
External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip
So Total net Force on each bolt Fb = P_Eb + Pi
Fb = 4.9 kip + 40 kip
Fb = 44.9 kip
Therefore the net force in every bolt is 44.9 kip
It is possible to generate a policy in which common points such as those mentioned above are agreed in order to hire or fire employees in their function of their psychological personality, that is, the character of knowledge and skills. Depending on the company, Test could be created in order to evaluate the psychological skills of the employees, as well as Test to periodically determine how their employees are kept up to date with regard to knowledge. The cumulative filter could be done every semester, for which each employee must exceed a minimum margin of score on these tests, otherwise his position could be at risk.
At the same time, incentives can be generated for the best scores that are rewarded not only with monetary values but also with rest days, coupons in restaurants or sports, which would cause the worker to strive to be constantly learning.
This policy agreement is outside the vision and mission of the company, and whose information must be given to the worker once he begins his work activities.
Answer:
μ = 0.136
Explanation:
given,
velocity of the car = 20 m/s
radius of the track = 300 m
mass of the car = 2000 kg
centrifugal force


F c = 2666. 67 N
F f= μ N
F f = μ m g
2666.67 = μ × 2000 × 9.8
μ = 0.136
so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136