Forces acting on two blocks of different masses and their relations (Newton's Laws)

Question

Could someone please help me with this question about Newton's laws and free body diagram?

"A block of mass M = 8.00 kg is located on a horizontal surface without friction. A second block of mass m = 2.00 kg is placed over the first. Neglecting possible friction between the contact surfaces, (a) what should be the acceleration value of the set (M + m) so that the smaller block does not slide over the larger one? (b) How is the force intensity F? Consider the slope of the block equal to 20°."

I really tried to draw the free-body diagram on each block, but I had a hard time relating the acting forces and their directions.

Answer 1

The force $F$ is accelerating the centre of mass of both blocks to the right with acceleration

$\displaystyle a = \frac {F} {M+m}$

Since the smaller block is stationary with respect to the larger one, it is also accelerating to the right with acceleration $a$, so

$N \sin 20^o = ma$

where $N$ is the normal force from the larger block. But the smaller block is in equilibrium vertically so

$ N \cos 20^o = mg$

Putting this together we have

$ a = g \tan 20^o \approx 3.57 \text{ ms}^{-2} \\ F = (M+m) a = 10a \approx 35.7 \text { N}$

Answer 2

If fricction can be disregarded, it means that the contact force $N$ between the two blocks will be orthogonal to their common interface (see figure). Furthermore, since the small block is not sliding, it means that it is not moving vertically, so the vertical resultant applied on it must be zero: $$N\cos{(20)} = mg$$ from which $N$ can already be obtained: $N \approx 20.88 \,\mathrm{N}$. Also, the horizontal component of the resultant is equal to $a$, the common acceleration of the two blocks, which are moving solidarly: $$N\sin{(20)} = ma$$ This yields $a \approx 3.57\,\mathrm{m}\mathrm{s}^{-2}$. Finally, the total force applied to the two masses together must fulfill $$F = (M+m)a \approx 35.7 \,\mathrm{N}$$