Answer:
Ff = 7.35*10^5 N
Explanation:
To find the maximum transverse pulling force you use the following formula, for the friction force:
[tex]F_f=\mu N=\mu Mg[/tex] (1)
μ: friction coefficient = 0.15
N: normal force, which is equal to the weight over the wheel
M: mass of th train = 500 ton = 500 000 kg
g: gravitational acceleration = 9.8 m/s^2
You replace the values of the variables in the equation (1):
[tex]F_f=(0.15)(500000kg)(9.8m/s^2)=735000 N=7.35*10^5N[/tex]
hence, the maximum transverse pulling force is 7.35*10^5 N
