Question

A system of three masses A,B and C shown in figure is pushed by a force F. All surfaces are smooth except B and C. Static frictional coefficient between B and C is m. Minimum value of F to prevent block B from downward slipping is

Answer 1

${\tt{\pink{ANSWER}}}$

$f \: = \frac{5mg}{2 μ}$

${\tt{\pink{Explanation}}}$

plz refer to the attachment above

Answer 2

(5/2μ)mg

Explanation:

Step 1: Drawing free body diagram [Ref. Fig.]

f→ Frictional force between B and C.

N 1 → Normal contact force between A and B

N 2 → Normal contact force between B and C

Step 2: Applying Newton’s second law

On all three blocks Combined:

From figure, we see that all three blocks will move with same acceleration(say a).

Therefore, Applying Newton's second law on the complete system(Blocks A, B & C combined)

(Positive Rightwards)

               ∑Fx = ma

                 F = (2m + m + 2m)a

                 F = 5ma

∴ Common Acceleration of the blocks, a= F/5m

On block C:

                N2 = 2ma

                      = 2m(F/5m)

                      = 2F/5

let the above equation be equation 1

Step 3: Condition of Minimum force

To prevent block B from slipping downward

                  f ≥ mBg  (mB = mass of block B)

let the above equation be equation 2

For minimum value of F, Maximum frictional force should act

So,            f max = μN2

                           = 2/5(μF)

let the above equation be equation 3

Step 4: Equation solving

Using equation (2) and (3)

                 2/5(μF) ≥ mg

                      F ≥ (5/2μ)mg

Hence minimum value of F to prevent block B from downward slipping is

(5/2μ)mg

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