Question

In the arrangement the coefficient of friction between the 2 blocks is`mu=1//2`What is the force of friction acting between the 2 blocks?

Answer 1

`8N`

Explanation:

For ,the two blocks to move together,we can say the frictional force acting at the interface of the two blocks must supply the required force(`ma=2a`) for the smaller block to move forward.

So,we can say for the larger block, `F-f=4a` (where,`a` is the net acceleration)

or, `f=F-4a=24-4a`

And,for the smaller block, `f=2a`

So,comparing both, `24-4a=2a`

or, `a=4 ms^-2`

So,`f=8N`

Let's,see maximum value of frictional force that can act at their interface ,it is `mumg=1/2*2*9.8=9.8N`

So,required frictional force is enough for the two blocks to move together.

Note,frictional force will not act by `9.8 N`,as the `2*a` amount of force is acting backwards relative to the block(thinking from non inertial frame of reference),so frictional force will act just by that amount,so that it will have no relative motion backward,i.e will go along with the lower block,if value of `ma` would have exceeded maximum value of frictional force,then the smaller block would have started moving backwards due to that extra amount of force.

Answer 2

`8` [N]

Explanation:

`m_1 = 2`
`m_2 = 4`
`mu = 1/2`
`g = 10`

Supposing no slipping the set is accelerated as

`(m_1+m_2) alpha_2 = F` or

`alpha_2 = F/(m_1+m_2)`

During carrying the upper block is submitted to a traction force of

`m_1 alpha_2`

this force compared with the maximum static traction force gives

`mu m_1 g < m_1 alpha_2 = m_1/(m_1+m_2) F`

so no slip occurs and the upper block is actuated with a force of `m_1/(m_1+m_2) F = 8`[N]

Answer 3

The force of friction acting on the 2 kg block will be 8 N in the direction of the 24 N force.

Explanation:

Analysis to determine whether or not the 2 kg block slips:

If slipping does not occur, what will be the acceleration of the blocks?

`F = 24 N = m*a = 6 kg*a`, therefore

`a = (24 N)/(6 kg) = 4 m/s^2`

Would the friction be able to support the 2 kg block accelerating at `4 m/s^2`?

The 2 kg block would need a force, `F_2 = 2 kg*4 m/s^2 = 8 N` to accelerate at `4 m/s^2`.
Will the force of friction, `F_f`, be able to provide 8 N?

`F_f = mu*N = mu*m*g`

`F_f = 1/2*2 kg*10 m/s^2 = 10 N`

Yes, `F_f` can, and will, provide the 8 N required for the 2 kg block to accelerate at `4 m/s^2`. So, the 2 kg block will not slip.

I hope this helps,
Steve