A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction $\frac{1}{\sqrt{3}}$. It is desired to make the body move by applying the minimum possible force F N. The value of $F$ will be ______ (Round off to Nearest Integer) [Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$]

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Laws of Motion

A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction

\frac{1}{3}

. It is desired to make the body move by applying the minimum possible force F N. The value of

F

will be ______ (Round off to Nearest Integer)
[Take

g = 10 ms^{- 2}

]

Answer: 5

Solution

A pull

F

applied at an angle

θ

above the horizontal lightens the contact,

N = m g - F sin θ

, and sliding is about to begin when

F cos θ = μ N

, so

F = \frac{μ m g}{cos θ + μ sin θ}

The denominator is largest, and

F

smallest, when

tan θ = μ

, where it equals

1 + μ^{2}

. With

μ = \frac{1}{3}

m = 1

kg and

g = 10 m s^{- 2}

F_{m i n} = \frac{μ m g}{1 + μ ^{2}} = \frac{10/ 3}{4/3} = \frac{10/ 3}{2/ 3} = 5

The smallest force that will start the body moving is

5

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Numerical answer

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