- #1

Sixth42

- 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.