 #1
 2
 0
Homework Statement
The attachment above contains the diagram given in the problem
M_{1} = 25kg
M_{2} = 35kg
Friction coefficient between M_{1} and stucture = 0.15
Friction coefficient between M_{1} and M_{2} = 0.25
Friction on pulleys and mass on pulleys is negligible.
Find the minimum mass M_{3} for sliding to occur in the system (i.e. the maximum mass for which the system will remain in static equilibrium).
Homework Equations
∑F_{x} = 0
ΣF_{y} = 0
The Attempt at a Solution
For the first mass:
The above attachment contains the diagram I used with the rotated coordinate system. My equations were:
∑F_{y} = 0
M_{2}g sin(60°)M_{1}g sin(60°) + Fn = 0
Fn = 509.7N
∑F_{x} = 0
M_{2}g cos(60°)M_{1}g cos(60°)  0.15Fn + T_{1} = 0
T_{1} = 370.8N
For the second mass:
The above attachment contains the diagram I used with the rotated coordinate system. My equations were:
∑F_{y} = 0
M_{2}g sin(60°) + Fn_{2} = 0
Fn_{2} = 297.3N
∑F_{x} = 0
M_{2}g cos(60°)  0.25Fn_{2}  T_{2} + T_{1} = 0
T_{2} = 124.8N
For the pulley attached to M3:
∑F_{y} = 0
2T_{2}  M_{3} = 0
M_{3} = 249.6N
Therefore mass of M_{3} = 249.6/9.81 = 25.44kg
I managed to get a final answer of M_{3} = 25.44kg (or 249.6N) by working on each mass individually. Using a rotated coordinate system for M_{1} and M_{2}. It would be nice if someone could confirm this answer for me but my main questions are:
Was I correct in adding M_{2} to the force diagram for M_{1}?
In both diagrams, is the friction force in the correct direction?
Does this method look correct overall or did I miss something?
Thanks for the help.
Attachments

8.4 KB Views: 409