Question

In the system shown in the figureM=13.4kg,m1=1kg,m2=2kg and m3=3.6kg the co efficient of static friction between m1 and m2 is 0.75 and that between m2 and M is0.6 All other surfaces are frictionless the minimum horizontal forceF must be applied to M sothat m2 doesn't slip on M is​

Answer 1

Explanation:

Given :

Let m1 = 1 kg

m2 = 2 kg and

m3 = 3 kg

Suppose the block m1 moves upward with acceleration a1 and the blocks m2 and m3 have relative acceleration a2 due to the difference of weight between them.

Actual acceleration of the blocks m1, m2 and m3 will be

a1, (a1 − a2) and (a1 + a2)

From figure 2, T − 1g − 1a1 = 0    …(i)

From figure 3,

T/2-2g-2(a1-a2)=0    …ii

From figure 4,

T/2-3g-3(a1+a2)=0    …iii

From equations (i) and (ii), eliminating T, we get:

1g + 1a2 = 4g + 4 (a1 + a2)

5a2 − 4a1 = 3g    …(iv)

From equations (ii) and (iii), we get:

2g + 2(a1 − a2) = 3g − 3 (a1 − a2)

5a1 + a2 = g    …(v)

Solving equations (iv) and (v), we get:

a1=2g/29

a2=g-5a1

a2=g-10g/29

=19g/29

Then  a1-a2=2g/29-19g/29

= -17g/ 29

and a1+a2=2g/29+19g/29

=21g/29

So, accelerations of m1, m2 and m3 are

19g/29up, 17g/29 down and 21g/29down, respectively.

Now, u = 0, s = 20 cm = 0.2 m

a2=19g/29        

 ∴ s=ut+1/2at

⇒0.2=1/2×19/29gt²

⇒t=0.25 s