Question

You are given that the coefficient of kinetic friction between the block and the table in Fig. 4.36 is 0.560, and m1 = 0.135 kg and m2 = 0.260 kg.

(a) What should m3 be if the system is to move with a constant speed?
(b) If m3 = 0.095 kg, what is the magnitude of the acceleration of the system? (Assume ideal conditions for the string and pulleys.)

Answer 1

The equation \[\Theta _{k }=\mu _{k}\times N\] is used for describing the relation between the normal force N between the two bodies, the mu is the kinetic friction coefficient and the theta is the resulting kinetic frictional force.
Now
\[\mu _{k}=0.560\]
answer a:
m2-m1 = 0.260-0.135=0.125. The force that m1 and m2 exert together on m3 is F = m.g (law Newton) = 0.125kg*9,81m/s² (g is the Gravitational acceleration) = 1.22625 kgm/s² = 1.22625 N (N stands for Newton).
This is theta. Now we can calculate N
N = Theta/µ = 2.1897 Newton
Mass of m3 is thus this required weight devided by the Gravitational acceleration:
m3 = N/g = 2.1897/9.81 = 223.21 gram
but you will have to give the construction an initial speed.

answer b:
m3 = 0.095kg
N =m3*g = 0.93195 N
Again
\[\Theta _{k }=\mu _{k}\times N\]
Theta = 0.521892 N
Total accelerating force is the netto force on m3 exerted by the two masses m2 and m1, minus the friction force from m3

Total force = 1.22625 N (calculated in answer a) - Theta
= 0.704358 Newton

Answer 2

Oops, still the acceleration: with m the total mass of all moving objects
F=m.a => a = F/m = F/(m1+m2+m3) = 0.704358 N / 0.49kg = 1.437 m/s²