Question

An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle θ with a plane, then the magnitude of the force is F = μW μ sin(θ) + cos(θ) where μ is a constant called the coefficient of friction. For what value of θ is F smallest?

Answer 1

Answer:$\theta =\arctan \mu$

Step-by-step explanation:

we know force sin component would oppose the weight  of object thus normal reaction will not be W rather it would be

$N=W-Fsin\theta$

therefore force cos component will balance the friction force

F$cos\theta$ =$\left ( \mu N\right )$

F$cos\theta$ =$\left ( \mu \left ( W-Fsin\theta \right )\right )$

F=$\frac{\mu W}{cos\theta +\mu sin\theta}$

F will be smallest when $cos\theta +\mu sin\theta$ will be maximum

and it will be maximum when we differentiate it to get

$\theta =\arctan \mu$