# Determin max aceleration of obj A200kg to ryt alng ground so obj B20kg ontop of A atached by rope ovr pully to obj C2kg hung ovr side of A wil nt slipall coefficients of static friction = 0.35....

Determin max aceleration of obj A200kg to ryt alng ground so obj B20kg ontop of A atached by rope ovr pully to obj C2kg hung ovr side of A wil nt slip

all coefficients of static friction = 0.35. Rope = no mass. Pully = no friction. Sorry for shortening but restricted amount of characters.

valentin68 | College Teacher | (Level 3) Associate Educator

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The free body diagram for masses B and C is below. The mass B does not slip on to the mass A means that they have the same acceleration. Also mass C is tied to mass B means they have the same acceleration.

`a_A = a_B =a`

`a_C =a_B (= a)`

When a force F will pull on the mass A , because of its inertia mass B will tend to remain behind A (although it is tied to C), thus the friction force on B is directed to the right (ahead).

Because B is sitting on A and not slipping, the static friction force on B is increasing linearly until a maximum force (when B begins to slip). This maximum friction force on B is

`Ff_B =mu*m_B*g`

For masses B and C we can write

`T +mu*m_B*g =m_B*a`

`T +m_C*a =m_C*g`

subtracting the two equations above we get

`mu*m_B*g -m_C*a =m_B*a -m_C*g`

`a*(m_B+m_C) =g*(mu*m_B +m_C)`

`a = g*(mu*m_B+m_C)/(m_B +m_C) = 9.81*(0.35*20+2)/(20+2) =4.013 m/s =4 m/s`

The maximum acceleration A can have before B is beginning to slip on it is 4 m/s.

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