# Three slabs are stacked vertically with the lowest slab being on wheels as shown below. An applied force of 44N acts horizontally on the bottom slab. How do you find the answers to the following questions?

## a. What is the acceleration of the system if the slabs do not slip? b. How much friction acts between the bottom and middle slab? c. What coefficient of friction is necessary to prevent slipping between the middle and bottom slab? d. How much friction is needed between the top and middle slab to prevent slipping?

Nov 6, 2017

a. The slabs do not slip, three blocks act as one.
Mass of system of blocks$= 2 + 4 + 5 = 11 k g$

Force applied $= 44 N$

Using Newtons Second Law of motion

$\vec{F} = m \vec{a}$

Inserting given values

$44 = 11 | \vec{a} |$
$\implies | \vec{a} | = \frac{44}{11} = 4 m {s}^{-} 2$,
direction being the direction of applied force.

b. Mass of top two blocks resting on the lowest block$= 2 + 4 = 6 k g$
Acceleration of top two blocks $a = 4 m {s}^{-} 2$
Force applied on top two blocks$= m a = 6 \times 4 = 24 N$
As the blocks do not slip, this must be equal to Force of friction between middle and lowest block.
$\therefore$Force of friction between middle and lowest block$= 24 N$

c. We know that

Normal Reaction $= \mu m g$

From Newton's Third Law of motion we know that Action and reaction are equal and opposite.

$\therefore$ Normal Reaction on the surface of bottom block$= 24 N$
Taking $g = 10 m {s}^{-} 2$, we get
$24 = \mu \times 6 \times 10$
$\implies \mu = \frac{24}{6 \times 10} = 0.4$

d. Action Force applied by top block on the middle block $= 2 \times 4 = 8 N$
As explained in b.

To prevent blocks from slipping friction required between top and middle blocks $= 8 N$