Question Number 102553 by aurpeyz last updated on 09/Jul/20

Commented byaurpeyz last updated on 09/Jul/20

pls help

Commented byaurpeyz last updated on 09/Jul/20

Commented byRio Michael last updated on 10/Jul/20

Force = μF_N  + mg sin θ  ⇒ F = (0.4)(10 × 9.8)N + (10)(9.8)sin 60N

Answered by mr W last updated on 10/Jul/20

F cos θ−mg sin θ≥μ(mg cos θ+F sin θ)  F (cos θ−μ sin θ)≥mg(μcos θ+sin θ)  ⇒F≥((mg(μ cos θ+sin θ))/(cos θ−μ sin θ))  ⇒F≥((mg(μ+tan θ))/(1−μ tan θ))  ⇒F≥((10×9.81(0.4+(√3)))/(1−0.4×(√3)))=681 N

Commented byaurpeyz last updated on 10/Jul/20

your solution is great Sir. pls explain the first line of the solution. i will understand the rest

Commented byaurpeyz last updated on 10/Jul/20

with this solution it seem like there is a force action at Θ degrees to the plane

Commented bymr W last updated on 10/Jul/20

Commented byaurpeyz last updated on 10/Jul/20

where F is the horizontal force right? so i will just have to resolve to get the answers like hou did

Commented byaurpeyz last updated on 10/Jul/20

Sir Pls check Q. 102194. what will be the direction of the friction?

Commented bymr W last updated on 10/Jul/20

in this question you should move  the object up to the plane and the  friction is always in the opposite  direction to the movement the object  tries to make. in this case it is  downwards.  but in Q102194 you don′t need to move  up the object, you just need to  prevent it to move down, in that case  the friction is upwards, because the  object tries to move downwards.

Commented byaurpeyz last updated on 10/Jul/20

i understand you Sir. thanks alot.

Commented byaurpeyz last updated on 11/Jul/20

Sir cam you recomend a textbook for me Sir. that will help me with all of this?

Commented bymr W last updated on 11/Jul/20

no sir. not because i don′t want to tell  you, but because i don′t know such a  book, not because i don′t find the  books good enough, but because i have  never read such a book since i left  the school.

Commented byaurpeyz last updated on 11/Jul/20

okay Sir