1) Ok so first lets draw a diagram.
Refer to image 1 below.
Friction will be up the plane since it needs to balance the force of gravity down the plane, the reason it needs to balance it is because it is in equlibrium.
We need to find r to find μ, so to do that we need to resolve the forces
perpendicular to the plane.R=mgcos(a) ....1
Next resolve the force parallel to the planeμR-mgsin(a)=0
so μR=mgsin(a) ....2
or μ=(mgsin(a))/R
Now substitute 1 into 2
μ=(mgsin(a))/mgcos(a)
so the mg's cancel and sin on cos is tan
so μ=tan(a)
Now finding acceleration.
So inclination is increased to b and the object is no longer in equilibrium. So we need to resolee the forces but this time letting the parallel force equal ma (i.e. Fnet)
Perpendicularso R=mgcos(b)
Paralleltake mgsin(b) first as it will be large as the acceleration is down the plane
mgsin(b)-μR=ma
mgsin(b)-μmgcos(b)=ma
so a=mg(sin(b)-μcos(b))/m
so then a=g(sin(b)-μcos(b)) m/s
2***Alternate solution Essential uses****Yeh they did it a silly way. this is what they did from the mgsin(b)-μ*R=ma part
-mgcos(b)\frac{sin(a)}{cos(a)}g=ma)
as μ=tan(a)
then
cos(a)-gcos(b)sin(a)}{cos(b)})
then use the sin(x-y)=sin(x)cos(y)-cos(x)sin(y)
to get a=g*sin(b-a)/cos(a)
2) We know that when the acceleration reachers g/4, the particle slips, in other words the maximum friction force has been exerted. So the Frmax=μR=the force under an acceleration of g/4
=mg/4
Now we need just μ so we need to find R and sub it in.
R will be mgcos(30)=mgroot(3)/2 N
Resolve parallel to surfaceμR-mgsin(30)=mg/4
μroot(3)/2mg-mg/2=mg/4
cancel the mg's
root(3)μ/2=3/4
μ=(3/2root(3))*root(3)/root(3) (rationalising)
μ=root(3)/2
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