Force and friction?

Question

A force F acts to the right on a 7.41 kg block. A 1.63 kg bloick is stacked on top of the 7.41 kg block and can slide on it witha coefficient of friection of 0.15 between the blocks. the tablee has a coefficient of friction of 0.2.
The acceleration of gravity is 9.8 m/s^2
The system is in equilibrium .
Find the force required to accelerate the 7.41 kg block at 2.6m/s^. Answer in units of N.

Answer 1

First, let's see if the 1.63 kg block will slip under that acceleration
F=m*a
must be less than
m*g*µ
or
a 9.8*.15
=1.47
2.6 is greater, so the top block will slip. I will solve the problem while the top block is in contact

The force of friction is
1.47*1.63
=2.40 N

Looking at a FBD of the lower block there are three forces acting on the block:

The external force to accelerate the block, the friction at the table, and the friction between the upper and lower blocks.

the normal force at the table is
(7.41+1.63)*9.8
=88.6 N
that makes the frictional force at the table
88.6*.2
=17.72 N

using F=m*a
F-17.72-2.4=7.41*2.6
=17.72 N

using F=m*a
F-17.72-2.4=7.41*2.6
F=17.72+2.4+7.41*2.6
F=39.4 N

j

Answer 2

interior the vector equation, F = ma, (F and a are vectors) F is the internet rigidity, the sum of all forces performing on the physique (utilized rigidity, gravity, friction, etc.) The frictional rigidity would be in a path opposing the action.