A body mass of 1 kg rests on a horizontal floor with which it has a coefficient of static friction. 31 .It is desired to make the body move by applying the minimum possible force.FN. The value of F will be _____________________ .(Round off the Nearest Integer)[ Take g=10 ms−2]

Name: Video solution 1: A body mass of 1 kg rests on a horizontal floor with which it has a coefficient of static friction. 31 .It is desired to make the body move by applying the minimum possible force.FN. The value of F will be _____________________ .(Round off the Nearest Integer)[ Take g=10 ms−2]
Uploaded: 2023-10-04T15:14:13.000Z
Duration: 3 min
Description: Video solution 1: A body mass of 1 kg rests on a horizontal floor with which it has a coefficient of static friction. 31 .It is desired to make the body move by applying the minimum possible force.FN. The value of F will be _____________________ .(Round off the Nearest Integer)[ Take g=10 ms−2]

tanϕ=NμsN=μssinϕ=μs2+1μs Block is in limiting equilibriumLami's theoremsin(π−ϕ)F=sin(90−θ+ϕ)mgF=cos(ϕ−θ)mgsinϕ when θ=ϕ Fmin=mgsinϕ=μs2+1mgμs=31+1mg31=5 N

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Equilibrium of forces

A body mass of 1 {~kg} rests on a horizontal floor with which

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Medium

Solving time: 1 mins

A body mass of $1 kg$ 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.
FN. The value of $F$ will be _____________________ .(Round off the Nearest Integer)
$[$ Take $g = 10 ms^{- 2}]$

Views: 5,499 students

Updated on: Oct 4, 2023

Exam:

JEE Mains 2021 Mar 17 Shift 2

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Noah Discussed

A body mass of

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.
FN. The value of

F

will be _____________________ .(Round off the Nearest Integer)

[

Take

g = 10 ms^{- 2}]

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

tan ϕ = \frac{μ _{s} N}{N} = μ_{s} sin ϕ = \frac{μ _{s}}{μ _{s}^{2} + 1}

Block is in limiting equilibrium
Lami's theorem

\frac{F}{sin ( π - ϕ )} = \frac{m g}{sin ( 90 - θ + ϕ )} F = \frac{m g sin ϕ}{cos ( ϕ - θ )}

when

θ = ϕ

F_{m i n} = mg sin ϕ = \frac{mg μ _{s}}{μ _{s}^{2} + 1} = \frac{mg \frac{1}{3}}{\frac{1}{3} + 1} = 5 N

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Question Text	A body mass of $1 kg$ 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. FN. The value of $F$ will be _____________________ .(Round off the Nearest Integer) $[$ Take $g = 10 ms^{- 2}]$
Updated On	Oct 4, 2023
Topic	Laws of motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	232
Avg. Video Duration	4 min

Text solutionVerified

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Text solutionVerified

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