Mechanics problem - Two masses and a pulley with working

Dec 2012
4
0

For mass y, I set my coordinate system to be the usual so:

W = -yg
T1 + W = 0, so T1 = yg

For mass x, I set my coordinate system so that +ve i is directed up the plane and +ve j is directed perpendicular to the plane (ie. the normal reaction)

N = Nj
F = Fi
W = -xg sin θ i - xg cos θ j
T2 = |T2| sin θj - |T2| cos θ i

I need to show that, to remain in equilibrium:

i) (mass) X >= Y tan θ

Since its a model pulled, the tension is the same on both sides. Hence, I figure this is something to do with the tension in the rope having to be less than the weight of X. Do I need to make the coordinate systems the same for both particles? If someone can point me in the right direction that would be helpful.

and ii) that μ >= (X sin θ + y cos θ) / (X cos θ - y sin θ)

My working so far is:

Resolving my equations in i gives |F| = |Xg| sin θ + |T2| cos θ

Resolving in j gives |N| = |Xg| cos θ + |T2| sin θ

And T1 = T2 = -yg (model pulley)

F <= u N

so u >= (Xg sin θ + yg cos θ) / (Xg cos θ + yg sin θ)

Obviously, we can divide each term by g and this is close, so I know Im going along the right lines, but the denominator should be (X cos θ - y sin θ) and Im not sure where my mistake is to make it a -ve.


Thanks for any help!
 

Unknown008

PHF Hall of Honor
Jun 2010
609
137
Mauritius
|N| = |Xg| cos θ + |T2| sin θ

That equation is not good. Remember that the force due to the rope is opposing the force due to the weight of X.

So that it should be: |N| = |Xg| cos θ - |T2| sin θ

(N is acting upwards and has to be positive, and the weight should be larger than the force due to the rope)


For the first part, simply resolve along i. What is the requirement for the system to remain in equilibrium along i only?
 
Dec 2012
4
0
Thanks for the help with (ii)

Did you mean resolve in j?

For the system to remain in Equilibrium, the resolution of j should be greater than or equal to 0 (ie. N >= 0 ) since if it was -ve the block would not be in contact with the plane.

Then |N| = |Xg| cos θ - |T2| sin θ

so X cos θ >= Y sin θ

and X >= Y tan θ

as required?
 

Unknown008

PHF Hall of Honor
Jun 2010
609
137
Mauritius
Oh, yes, I thought i was the direction perpendicular to the plane for some reason :eek: