Question

Solve this ....................

Answer 1

(a) see the diagram ,
N₁ = normal reaction between surface and M ,
N₂ = normal reaction between m₁ and M
fr = friction between m₁ and M
same string use in both pulley so, Tension in each have same e.g T

(b) maximum value of fr₁ { friction b/w m₁ and M} = μm₁g = 0.3 × 20 × 10 = 60N

maximum value of f₂ = 0.3 × m2 × g
= 0.3 × 5 × 10 = 15 N
Forces on m₁and m₂ in horizontal directions have two possibilities :
(1)either m₁and m₂ will remain stationary or
(2) m₁and m₂ both are moving

for 1st case :-
T ≤ f₁ or T ≤ 60N
T≤ f₂ or T ≤ 15N
take common then T ≤15 N
when, we take T = 14N ,
for equilibrium,
T = f₁ = 14N,
T = f₂ = 14N
then,
f₁ = f₂= 14N
but a/c to question ,
f₁ = 2f₂
so, 1st case is not possible .

it means m₁and m₂ both are not stationary.

in 2nd case :- when m₁and m₂ are moving .
f₂ = 15 N { from above
f₁ = 2f₂
f₁ = 30 N
but maximum value of f₁ = 60N ,
hence, it means there's not realtive motion between m₁ and M .
it means acceleration is same for all .
Let acceleration = a

now,
for m1 :-
30 - T = 20a ______________,(1)
for m2
T - 15 = 5a ___________(2)
solve both equations ,
15 = 25a
a = 3/5 m/s²
now,
For M ,
F - f1 = Ma
F - 30 = 50a
F - 30 = 50 × 3/5 = 30
F = 60 N

hence, F = 60N, a = 3/5 m/s²
T = 5× 3/5 + 15 = 18 N