figure shows two blocks connected by a light string placed on the two inclined parts of a triangular structure. The coefficients of static and inetic friction are 0.28 and 0.25 respectively at each of the surface. a.Find the minimum and maximum values of m for which the system remains at rest, b. Find the acceleration of either block if m is given the minimu value calculated in teh first part and is gentluy pushed up the incline for a short while.

Explanation

a. Take the 2 kg block as the system. The forces on this block are shown in ure with `M=2` kg. It is assumed that m has its minimum value so that the 2 kg block has a tendency to slip down.As the block is in equilibrium, the resutant force shoud be zero.


Taking coponents `_|_` to the incline
`N=Mg cos 45^0=Mg /sqrt2`.
Taking components || to the incline
`T+f=Mg sin 45^0 = Mg /sqrt2`. or ` T=Mgsqrt2-f`
As it is a case of limiting equilibrium.
`f=mu_2N`
or `T=Mg/sqrt2-mu_s Mg/sqrt2=Mg/sqrt(1-mu_2)`............i
Now consider the other block as the system. the forces acting on this block are shown in ure.
Taking components `_|_` to the incline,
`N'=mg cos 45^0+f'=mg/sqrt2+f'`
As it is the case of limiting equilirium
`f'=mu_sN'=mu_s mg/sqrt2`
Thus, `T=mg/sqrt2(1+mu_s)`
From i. and ii
`m(1+mu_s)=M(1-mu_s)`
or `m=(1-mu_s)/(1+mu_s) M=(1-0.28)/(1+0.28)xx 2 kg`
when maximum possible value of m is suplied the directions of friction are reversed becauyse m hs the tendency to sip down and 2 kg block to slip up. Thus, themaximum value of m can be obtained from iii by utting `mu=-0.28. Thus the maximum value of m is `brgt `m=(1+0.28)/(1-0.28)xx2 kg
`32/9kg`
b. If `m=9/8 kg ` and the system is gently pushed, kinetic friction will operate. Thus,
`f=mu_k Mg/sqrt2 and f'(mu_kmg)/sqrt2
where `mu_k=0.25`. If the acceleration is a Newton's second law for M gives ure.
`Mg sin 45^0-T-f=Ma`
or, `Mg/sqrt2-T-(mu_kMg)/sqrt2Ma`............iv
Applying Newton'ssecond law m ure.
`T-mg sin 45^0 -f' =Ma`
`T-Mg/sqrt2-(mu_kmkg)/sqrt2=Ma`a...........v
Adding iv and v
`Mg/sqrt2(1-mu_k)mg/sqrt2(1+mu_k)=(M+m)a`
or, `a=(M(1-mu_k)-m(1+mu_k))/(sqrt(2(M+m)) g`
`=(2xx0.75-9/8xx1.25)/(sqrt2(2+9/8))g`
`=031 m/s^2`.