Answer:
symbolic solution for the coefficient of friction between the baby and the sled is
[tex] \mu = \frac{m + M_s}{m} * \frac{a}{g}[/tex]
Explanation:
From the question we are told that
The mass of the boy is [tex]M_e =[/tex]
The mass of the shed is [tex]M_s[/tex]
The mass of the baby is [tex]m[/tex]
The maximum acceleration for the baby to stay is a
Generally the frictional force between the baby and the shed is mathematically represented as
[tex]F_f = \mu N[/tex]
Here N is the normal force acting on the baby which is mathematically represented as
[tex]N = m * g[/tex]
=> [tex]F_f = \mu * m * g[/tex]
Now for the baby not the slip off the shed the frictional force between the baby and the shed must be equal to the the force applied by the child
The force applied by the child is mathematically represented as
[tex]F = [M_s+m] * a[/tex]
So
[tex] \mu * m * g = [M_s+m] * a[/tex]
=> [tex] \mu = \frac{m + M_s}{m} * \frac{a}{g}[/tex]
