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# 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?

Grade:12

## 3 Answers

Rohith Gandhi
24 Points
11 years ago

Dear Eshita,

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.

All the best Eshita !!!

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Rohith Gandhi P

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

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)

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

All the best.

Regards,
Askiitians Experts
Badiuddin

Rishi Sharma
askIITians Faculty 646 Points
9 months ago
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

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