Laws of Motion 3 Question 21

22. A block of mass m rests on a horizontal floor with which it has a coefficient of static friction μ. It is desired to make the body move by applying the minimum possible force F. Find the magnitude of F and the direction in which it has to be applied.

(1987, 7M)

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Answer:

Correct Answer: 22. mgsinθ,θ=tan1(μ) from horizontal

Solution:

  1. Let F be applied at angle θ as shown in figure.

Normal reaction in this case will be,

N=mgFsinθ

The limiting friction is therefore

fL=μN=μ(mgFsinθ)

For the block to move,

Fcosθ=fL=μ(mgFsinθ) or F=μmgcosθ+μsinθ

For F to be minimum, denominator should be maximum.

 or ddθ(cosθ+μsinθ)=0 or sinθ+μcosθ=0 or tanθ=μ or θ=tan1(μ)

Substituting this value of θ in Eq. (i), we get

Fmin=mgsinθ



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