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If a particle on a rough inclined is attached to a spring so that the spring is parallel to the inclined place, when the particle is equilbrium just because of the weight of the particle extends the spring somewhat. Does friction need to be taken into account (because it is not moving), will it be the maximum friction $\mu R$ (coefficient of friction times the reaction)? and will it be acting up or down the slope?

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There is static friction even when the particle is not moving. Friction always acts in the direction to oppose motion. –  leongz Jun 9 '13 at 19:11
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Indeed we still need to consider friction, for if there was no friction the equilibrium of the spring would be extended even farther down the ramp. The friction we are interested is called static friction, which is contrasted with kinetic friction.

The force of friction would point up the ramp, since the static friction is keeping the ball from rolling down the ramp (along with the spring). There we see that the friction does some of the "work" in keeping the ball up on the incline, and the spring does the rest. If there was no friction the spring would need to do all the "work" of keeping the ball from rolling down the incline. I use work in quotation marks because there is no actual displacement and I meant work in the common sense (effort).

The actual force of the friction will be given by the coefficient of static friction, which is some constant, times the Normal Force. The normal force is the force the incline exerts on the ball, normal to the inclines surface: $F_N=mg\cos(\theta)$. Where $m$ is the mass, $g$ is the gravitational acceleration and $\theta$ is the angle of the incline. And thus the force of static friction would be: $F_{\mu_s}= \mu_s*F_N= \mu_s*mg\cos(\theta)$. Where $\mu_s$ is the static coefficient of friction.

Note: I believe the above has answered your question in full, but what follows is hopefully some justification for why the friction is important. It might be helpful to balance all the forces acting on the ball, so let's think about what those are.

i) gravity pull the ball down the ramp,

ii) the spring pulling the ball up the ramp, and

iii) the static friction holding the ball form rolling down the ramp.

Let's impose the condition that the ball is not moving so therefore $F_{net}=0$. Now let's find out what the net force looks like in terms of the three forces:

i)Gravity: $F_g=mg\sin(\theta)$ (Down the ramp)

ii)Spring: $F_{spring}=k\Delta x$ (Up the ramp). Where is $k$ is the spring constant and $\Delta x$ is the displacement from equilibrium.

iii)Friction: $F_{fric}=\mu_s*mg\cos(\theta)$ (up the ramp)

So that: $F_{net}=F_g+F_{spring}+F_{fric}=-mg\sin(\theta)+mg\cos(\theta)+k\Delta x=0$. Then we can solve for the displacement of the spring by solving for x, which follows that:

$\displaystyle \Delta x=\frac{mg}{k}(\sin(\theta)-\cos(\theta))$

Note that in the absence of friction, x would be larger since the friction contributes the $-\cos(\theta)$ term. Hence I have shown that friction indeed decreases the displacement of the ball attached to the spring, ergo we do need to take friction into consideration!

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Thanks. Though you say friction acts up the ramp because the ball would roll down the ramp due to gravity. How do you know the ball wouldn't roll up the ramp due to the spring, and so friction should act down the ramp? –  Jonathan. Jun 10 '13 at 6:42
    
@Jonathan. ,if the spring is stretched to a point where the spring force exceeds the gravitational pull, the friction will act down the plane. Remember that friction always tries to oppose relative motion or the tendency of relative motion. Usually, we can calculate the net force due to all other forces except, and say that friction will act opposite to the net force vector. But beware, this only works if all the other forces are independent of the friction force. –  udiboy Jul 7 '13 at 6:53
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