1. Jan 17, 2006

### QueenFisher

i am feeling very stupid. i have done 3 virtually identical rough-inclined-plane-friction questions and they are all wrong.

mass 5kg on a rough inclined plane inclined at 30 degrees to the horizontal. is in limiting equilibrium. force F acting at 10 degrees from the parallel to the slope (vertically).

so. resolving parallel to the slope, Xcos10-5gcos60=F
F=(mu)R and mu=1/7
Xcos 10-5gcos60=1/7(5gcos30-xcos80)
and solving for X i get X=30.3

the other 2 are the same kind of thing. is there something fundamental that i'm getting wrong??

2. Jan 17, 2006

### finchie_88

Assuming that I understood the question...
What exactly does 'X' represent? and the frictional force is equal to the normal reaction perpendicular to the plane times by the friction constant, not (mu)R. So, R = mgcos30 - the vertical component of the other force acting at 10 degrees (assuming that that force is acting away from the plane). parallel components to the slope should mean that if its in equilibrium, mgsin30 = horizontal component of the other force acting at 10 degrees/ any other forces on the mass. I'm sure you can work it out from there.

btw, are you sure that mu = 1/7, that means an unusually small friction co-efficient of 1/35.

Last edited by a moderator: Jan 17, 2006
3. Jan 18, 2006

### QueenFisher

i meaunt mu as in that weird greek letter thingy (as in coefficient of friction = 1/7)
cheers for the help i think i can work it out now.