JEE Main Physics Class 11 Laws Of Motion Part 1 Questions

Let the force F is applied at an angle θ with the horizontal. For horizontal equilibrium,
F‌cos‌θ=µR .......(i)
For vertical equilibrium,
R+F‌sin‌θ=mg
or, R=mg−F‌sin‌θ .......(ii)
Substituting this value of R in eq. (i), we get
F‌cos‌θ=µ(mg−F‌sin‌θ)
=µmg−µF‌sin‌θ
or, F(cos‌θ+µ‌sin‌θ)=µmg
or, F=

µmg

cos‌θ+µ‌sin‌θ

.......(iii)
For F to be minimum, the denominator (cos‌θ+µ‌sin‌θ) should be maximum.
∴

d

dθ

(cos‌θ+µ‌sin‌θ)=0
or ,−sin‌θ+µ‌cos‌θ=0
or, tan‌θ=µ
or, θ=tan−1(µ)
Then, sin‌θ=

µ

√1+µ2

and
cos‌θ=

1

√1+µ2

Hence, Fmin=

µw

1 √1+µ2 + µ2 √1+µ2

=

µw

√1+µ2