Question

A system of masses is shown in the figure. calculate:
(i) the maximum value of F for which there is no slipping anywhere. (And=90N)
(ii) the minimum value of F for which B slides on C. (Ans=112.5N)
iii) the minimum value of F for which A slips on B(And=150N).


P.S:
pls solve the question along with explanations.( pls give the answer on a paper)
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Answer 1

Answer:

1. No slipping anywhere

Whole system moves with acceleration a, which is given by,  

a = (F-μN) / (20+30+40) = (F-88.2) / 90 m/s^2

∵ μ is the friction coefficient between ground and block-A.

N is the normal reaction force between the surfaces of ground and block-A

From the free body diagram given for block-A, we get the no slip condition for block-A is ( mA × a )< f ,  where f is frictional force between the surfaces of A and B.

$m_A * a < \mu *m_A *g$

We can write (F-88.2) < 90×μ×g  or ( F-88.2) < 90×0.1×9.8  or F < 176.4 N

Hence if the applied force F is less than 176.4 N, there will not be any slip between blocks.

 

2. Block-B sliding on Block-C

Free body diagram of block-B is given in figure.

We can see that if applied force exceeds sum of firctional forces f1 and f2 , block-B slides on block-C

 F >(f1 + f2) = (0.1×20 + 0.2×50)×9.8 = 117.6 N

 

3. Block-A slips on Block-B

As we already calculated for 'no slip anywhere', block-A slips if applied force exceeds 176.4 N

Answer 2

Explanation:

Whole system moves with acceleration a

which is given by

a=

20+30+40

(F−uN)

=(90 F−88.)ms −2

........(1)

u=coefficient of friction between A and ground

N= Normal rxn force b/w ground and A

For no slip condition of block A

m

A

a<F [F=frictionless force b/w A and B]

ma<m

A

g...........(2)

using eq (1) and eq (2)

F−88.2<umg⇒F−88.2<90×0.1×98

⇒F−88.2<88.2⇒F≤176.4N.

Hope it helps u.