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  • A force F is applied on a block of mass 5kg placed on an inclined plane of angle 300 as shown in the diagram. If the coefficient of friction between the block and plane is <img

A force F is applied on a block of mass 5kg placed on an inclined plane of angle 300 as shown in the diagram. If the coefficient of friction between the block and plane is \mu_{s}= 0.8 What is the minimum force (in N) to make the block down the inclne? (g= 10m/s2)

Option: 1

7.6


Option: 2

0.5


Option: 3

10


Option: 4

4.7


Answers (1)

best_answer

Given-

The mass of the block, m=5kg

Coefficient of friction, \mu= 0.8

Let us assume the block is pulled applying force F making an angle \alpha with the downward incline. F is the minimum force required to move the block downward. At this stage the block is just begining to move therefore limiting friction will be acting upward and block is in equilibrium.

F.B.D of the block-

Along the plane of incline-

\\ f_{l}=Fcos\alpha+mgsin30^0\ ...(1) \\ \mathrm{ perpendicular\ to\ the\ incline- }\\ \mathrm{Fsin\alpha+N=mgcos30^0 }....(2)\\

\\ \mathrm{From\ equation\ (1) and\ (2)- }\\ \mathrm{ Fcos\alpha+\frac{mg}{2}=\mu(\frac{\sqrt{3}mg}{2}-Fsin\alpha)\ }\\ \mathrm{Fcos\alpha+\mu Fsin\alpha=\frac{\mu\sqrt{3}mg}{2}-\frac{mg}{2}\ }\\ \mathrm{ F=\frac{ (\mu \sqrt{ 3}-1)mg}{2(cos\alpha+ \mu sin\alpha )}}

As numerator of the term is constant, for minimum value of F, denominator of the term must be maximum.

Given

\\ \mathrm{ (acos\theta+bsin\theta)_{max}=\sqrt{a^2+b^2}}\\

\\ \mathrm{ \Rightarrow F_{min}=\frac{ (\mu \sqrt{ 3}-1)mg}{2\sqrt{\mu^2+1}}=7.6N}\\

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Sayak

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