Question
A block of mass m is at rest on a rough inclined plane of angle of inclination θ . If coefficient of friction between the block and the inclined plane is required to move the block on the plane is
how the answer is mg[sinθ - μ cosθ] .
i think one line is missing in the question.
Read 2 Solution.
I think something u have skipped in the question
if u r asking for the force F required to move the block then d first step is to resolve mg in its two component & then make balance condition R=mgcosθ & F= mgsinθ + f f=uR
F= mg sinθ + µR =mg sinθ+ µmg cosθ =mg( sinθ+cosθ) as d body is moving up therefore friction f will act opposite to applied force in the direction of mgsinθ thus sign in b/w should be + , whereas u have used -ve
if u asking to move the body up then the solution submitted will be correct, But if u are asking for just to make it move or to move downward then the sign vyl be changed then balance conditn will be
F = mgsinθ-f = mg (sin θ -ucosθ) here f=uR & r=mgcosθ as force to move it is along sin component thus sign will be same 7 as friction f is acting opposite thus it has -ve sign . thus both ans are applicable depending upon the force is req in which direction