Two rectangular blocks are stacked on a table. The upper block has a mass of 4.9 kg, the lower block has a mas

Question

Two rectangular blocks are stacked on a table. The upper block has a mass of 4.9 kg, the lower block has a mass of 27.1 kg. The coefficient of kinetic friction between the lower block and the table is 0.3. The coefficient of static friction between the upper and the lower blocks is 0.605. A string is attached to the lower block, and an external force is applied in horizontal direction to pull on the string, as shown. What is the maximum force that one can apply without having the upper block slide off?

Answer 1

Interesting problem

The block on top with mass m1 must be subjects to a force equal that of friction between it and the lower block.
F1=f1
f1=uN1=umg
F1=m1a1
so
m1a1=u1m1g
a1=u1g
If m1 accelerated in one direction that means m2 is accelerated with the same magnitude but in apposite direction (so says the concept of inertia). Than means a1=-a2
for block #2 th eforce pulling it is the sum of its inertia plus the force of friction.
Ft=F2+f2 (force on the lower block plus the force of friction)

f2=u2(m1+m2)g
F2=m2a2=m2a1


Ft=m2a1+u2(m1+m2)g
Ft=m2u1g+u2(m1+m2)g
Ft=g(m2u1+u2(m1+m2))
Ft=9.81( 27.1 x .605 + .3(4.9 +27.1))
Ft=255.0 N
That is it

Answer 2

there became one man or woman, I gave the link to a splash, although the alternative became his. i did not outright ask him to dam that man or woman. from time to time trolls will objective your peeps and that is not cool. xo