Question

detailed explanation plsssxxxx​

Answer 1

Answer:

F(cos alpha +usin alpha) +mg(sin alpha-u cos alpha)

Explanation:

Resolve F along the incline you get Fcos alpha and Fsin alpha perpendicular to the incline

And also the component of weight which is mgsin alpha also acts down the incline

As the box moves downward frictional force which is uN acts up the incline

As N+Fsin alpha=mgcos alpha

N=mgcos alpha-Fsin alpha

So frictional force is u(mgcos alpha-Fsin alpha)

So the minimum force required will be

(Fcos alpha+mgsin alpha)-(umgcos alpha-uFsin alpha)

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Hope it helps

Answer 2

Answer:

syhffh ghkkhghj was tum of the 6th

Explanation:

plz tell you what we are doing to help with