How do I find the maximum value of the force between two blocks?

Question

The problem is as follows:

In the figure from below there are two blocks, one over another. The system is at rest. The horizontal surface is frictionless and the coefficient of static friction is $0.45$ Find the maximum value of $F$ in $N$ such as the blocks will not slide between them. (You may use $g=10\frac{m}{s^{2}}$)

The alternatives given in my book are:

$\begin{array}{ll} 1.&58\,N\\ 2.&42\,N\\ 3.&30\,N\\ 4.&25\,N\\ 5.&10\,N\\ \end{array}$

In my attempt to solve this problem I thought that:

For the lighter block:

$F-mg\times \mu_{s}=0$

$F= 5\times 10 \left(0.45\right)=22.5\,N$

But this ain't the case. I don't know how to relate this with what is happening in the block from below.

The answer supposedly is $30\,N$. But I have no idea how to get there. Can somebody offer some help here please?.

Answer 1

The maximum that can be applied to the lower block is $mg\mu_s=22.5N$. Thus the acceleration of the lower block would be $22.5/15=1.5m/s^2$

So for the blocks to not slide between them, the upper block can move at maximum $1.5m/^2$ acceleration. Which means $$F=mg\mu_s+m\times 1.5=30N$$

Answer 2

Your $\ 22.5\ N\ $ is the maximum horizontal force that can be transferred by friction from the $\ 5 kg\ $ block to the $\ 25\ kg\ $ block without any slippage occurring. Subject to that horizontal force the lower block will accelerate at $\ a=\frac{22.5}{15}\ m/s^2\ $. For the $\ 5\ kg\ $ block to accelerate at the same rate, the nett horizontal force on it, which is $\ F-22.5\ N\ $, must be $\ \frac{5\times 22.5}{15}=7.5 N\ $. Therefore $\ F=22.5+7.5=30\ N\ $.

Answer 3

Attempt:

Small body :$m$; Big body : $3m$.

1) Both are moving with the same acceleration $a$:

Only an external force $F$ is acting on the system :

1) $F=ma +3ma=4ma$;

2) Maximal force due to friction pulling body $3m$ with maximal acceleration $a_m$:

$3ma_m=\mu m g$;

$a_m= (1/3)\mu g$;

3) $F_m=4ma_m=$

$4 (5)(1/3)(0.45)10 [N];$

$F=20\cdot 1,5 [N] =30 [N]$.

Answer 4

Hint.

For the small body

$$ F -\mu m_1 g = m_1 a $$

for the large body

$$ \mu m_1 g = m_2 a $$

(The two bodies move at the same acceleration)

then

$$ \frac{1}{m_1}\left(F-\mu m_1 g\right) = \frac{1}{m_2}\mu m_1 g $$