a 5kg block rest on the 30degree incline.

Answered question

2022-02-18

a 5kg block rest on the 30degree incline. the coefficient of static friction between the block and the incline is 0.2 . how large a holizontal force must push on the block if the block is to be on the verge of sliding (a) up the incline ,(b) down the incline

Answer & Explanation

Vasquez

Vasquez

Expert2022-03-12Added 669 answers

if the horizantal force on the block =F

the component of the horizantal force in thedirecti on normal to the surface =Fsinθ

and the horizantal component of the force up ward the inclined plane =Fcosθ

the net normal force on the block =N=mgcosθ+(Fsinθ))

and the frictional force on the block =Ff=μsN

μs coefficient of static friction between block and ramp =0.20

the gravitational force component down the incline =mgsinθ

if the block stay at the verge of sliding up. the force balance equation is
Ff+mgsinθ=Fcosθμs[mgcosθ+(Fsinθ)]+mgsinθ=Fcosθmgcosθμs+(Fsinθμs)+mgsinθ=FcosθF=mgcosθμs+mgsinθcosθsinθμs

mass of the block =m=5.0 kg

g=9.8 m/s2;  θ=30;  μs=0.2

we plug the know values in the relation gives

F=43.06 N

a=0 mean the net force on the block is Fnet=0;

which physically will not bring any change in the position of the block.

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