A heavy body of mass 10 kg is to be dragged along a horizontal plane(μ=1/√3).The least force required is
106Let the force be F and angle with horizontal be θ.
Then N = mg - Fsinθ
and Fcosθ = μN = μmg - μFsinθ
==> F = μmg/(cosθ+μsinθ)
differentiate F wrt θ,
μ = tanθ for which F is least
putting the above result in the expression for F,
Fmin = mgsinθ = mg*μ/√1+μ2
1but from where did the angle came?
it is dragged, so the angle should not be 0?????????
1here the question is asked for the least for required to pull the block.
u should note that the least force is required not when it is pulled horizontally but when pulled at an angle as shown in the diagram.
also note that force of friction is directly proportional to normal reaction.
when force is applied at an angle, the vertical component reduces the normal reaction and horizontal componont produces acceleration.
working is shown by ashish in post#2
1kya funu question hai yaar hats of to u ek word ne pure answer ko badal diya.
1the angle of friction fro minimum force is tan inverse(1/U)
