FRICTION ?

[monkeywantsabanana]

Hey guys ! Can someone verify the answer to this question please?

A body of a mass 25kg rests in limiting equilibrium on a plane inclined at 30 degrees to the horizontal. If the plane is raised unit it is at an angle of 60 degrees to the horizontal, find:

a) the least force acting up the plane required to keep the mass in equilibrium

I've made a mistake somewhere...
I may have used wrong formulae? R=ma and F=µN ?

:-S what's the difference?

There's an attachment!

[xZero]

shouldnt it just be F=25gsin30 since you're trying to balance the downward force from gravity

[monkeywantsabanana]

sorry i'm not following.

[b^3]

ok so resolve forces up the plane
Fr-mgsin(angle)=ma
Fr-25gsin(60)=0
Fr=25root(3)/2 N
i.e. the force acting up the plane (that is including the friction) is 25root(3)/2.
If the answer is still wrong then minus the friction (the value you found) from this and this will give the force you need to apply up the plane to keep the object in equilibrium.
The equation would then be F+Fr-mgsin(angle)=0
where Fr=uN and F is the force you need to apply.

EDIT: Just to add to the understanding part of this. At the point it will slip, friction is a maximum i.e. Fr=uN (or uR). But friction may not be enough to hold it on the slope so another force is required (up the plane) to keep it there.

[monkeywantsabanana]

So the Force pulling the object up does not equal to the Friction? I'm still confused. :-S

[xZero]

yeah misread the question, basically in the 60 degree slope, friction force alone will not be enough to keep the object from sliding down, we need another force pushing it up. Since the force from gravity is 25root(3)/2 N, we also need an total upward force 25root(3)/2 N to balance it out. Part of this will be contributed by friction and the rest will be the least force required to keep the mass in equilibrium

[monkeywantsabanana]

ah ok, i kind of understand now but then i get F = -49.07 N can you have a negative force?

[b^3]

negative means it would be in the opposite direction. Did you make the forces up the plane positive or forces down the plane positive?

[monkeywantsabanana]

Well, i found friction in the attachment to be 25g(sqrt3)/6

Also figured out that F = 25g(sqrt3)/2 that's the force up the plane

So i did F - Fr = -49.07N

:-S it feels wrong...

Sorry & thanks for putting up with my idiocy.

[b^3]

ok lets put it all in the one equation
let say q = force required
Fr=friction
F= total force up the plane
q+Fr=F
so F-25gcos(60)=0
q+Fr-25g/2=0
q=25g/2-Fr
q=25g/2-uN
q=25g/2-25groot(3)/6
q=122.5-70
q=52.5N up the plane.

Note: friction will be up the plane since the object is trying to move down the plane.

[monkeywantsabanana]

dude, you're the best ! thanks so much !  makes sense, finally !

Cheers !

[b^3]

No problem. And plus this is helping me revise for spesh too.