A block of mass m is kept on a rough horizontal surface. The coeff. of static friction b/w the block and the surface is μ. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block? In which direction should this force act?

Dear Eshita,

 

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Rohith Gandhi P

Dear Eshita

 

F cos θ =μ N

           =μ(mg-Fsinθ)

F= μmg/(cos θ +μsinθ)

For minimum force  (cos θ +μsinθ) should be maximum

and maximum value of (cos θ +μsinθ) is √(1+μ²)

so F min =μmg/√(1+μ²)

and for the direction of force let

(cos θ +μsinθ) =√(1+μ²)(1/√(1+μ²)  cos θ  +  μ/√(1+μ²)  sinθ)

                     =√(1+μ²)sin(θ +φ)   wher tanφ =1/μ

 so sin(θ +φ)  =1

θ +φ =pi/2

θ =pi/2 -φ

    = pi/2 -tan^-1(1/μ)

     =tan^-1(μ)

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Askiitians Experts
Badiuddin

 

Dear Student,
Please find below the solution to your problem.

Let P be the force applied to it at an angle θ
From the free body diagram "
R+Psinθ-mg=0
R=-PSinθ+mg
mR=Pcosθ
From equation (i), μ(Psinθ-Pcosθ)
P=μmg/μSinθ+cosθ
Applied force should be minimum when msinθ+cosθ is maximum.

Again,
(μsinθ+cosθ) is maximum when its derivative is zero.

(d/dθ)(μsinθ+cosθ)=0
μCosθ-sinθ=0
θ=tan⁻¹μ\
So, P=μmg/μSinθ+cosθ
=μmglCosθ/(μsinθ+cosθ)/cosθ
By dividing the numerator and denominator by cosθ, we get
P=μmgsecθ/1+μtanθ
=μmgsecθ/1+tan²θ
=μmg/secθ
=μmg/√1+tan²θ
=μmg/(1+μ²)
Hence minimum force is μmg/√1+u² at an angle q=tan⁻¹μ

Thanks and Regards