66F_\mathrm{min}=\dfrac{\mu mg}{\sqrt{1+\mu^2}}
It should be applied at an θ with the horizontal such that \tan\theta = \mu.
A particle 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 direction in which it has to be applied.
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2 Answers
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let f= limiting frictional force=μmg
if F is applied parallel to the ground, then it is quite obvious that F≥μmg for the block to move
if F is aplied at an angle θ to the horizontal, then we draw the FBD and balancing the horizontal and vertical components we get
N= mg - F sinθ and
F=( μmg)/cos θ +μsin θ......for the condition when the external applied force just overcomes limiting fric force!
for min F , den must be max,,,,after suitable maximizations and minimizations , we get FMIN =(μmg)/√(μ2+1)
that is the answer !! job done !
