#1
This question is seriously pissing me off. I dont understand where I am getting wrong with it.Probably it is something really simple but I cant figure it out. Here i the question:

A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a horizontal force X. The coefficient of friction between body and plane is 0.2 . Find X when the body is on the point of slipping.

a)up the plane
b) down the plane

So I managed to do some of it

R-2gcos20=0
R=2gcos20

F= mu. R

F= 0,2 . 2gcos20

Then

I suppose if its going to slip up the plane then :

X-F-2gsin20=0

X= 0,2 . 2gcos20 + 2gsin20
X= 10.39N (but the answer in the book is 11.9N)??

If it is slipping down the plane then F>X ... and the how the forces will be arranged?
Thanks for the help in advance. And rep for everybody who doe help !
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12 years ago
#2
X is horizontal, not directed along the plane.

So, e.g., for part a:
X cos20 = 2g sin20 + f = 2g sin20 + 0.2 (2g cos20 + X sin20)

So we find that X=~11.9N.
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#3
still dont understand it though . How did you get to such equation ?
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#4
still dont understand it though . How did you get to such equation ?
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12 years ago
#5
I hope this attachment makes it more clear.
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#6
All I see ! I will try to figure it out now. Thanks a lot for the effort. I greately appreciate it . My trength is pure maths and I am quite weak at mechanics.
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