Solving a Static Equilibrium Problem - Help Appreciated!

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SUMMARY

This discussion focuses on solving a static equilibrium problem, specifically calculating the center of mass for a system involving a bar and a crate. Key steps include selecting a pivot point, determining the center of mass for both objects using their mass distributions, and applying the center of mass formula for both x and y components. The final calculated positions are approximately x_{cm} = 1.28m and y_{cm} = 0.67m from the pivot point. Additionally, the discussion emphasizes the application of Newton's first law to solve for forces and torques in the system.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Familiarity with center of mass calculations
  • Basic knowledge of trigonometry (sine and cosine functions)
  • Application of Newton's first law of motion
NEXT STEPS
  • Study the concept of center of mass in more complex systems
  • Learn about static equilibrium problems in physics
  • Explore the application of Newton's laws in various scenarios
  • Practice solving problems involving forces and torques
USEFUL FOR

Students in introductory physics courses, particularly those studying mechanics and static equilibrium, as well as educators looking for examples to illustrate these concepts.

Mark_iv
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Homework Statement



http://img76.imageshack.us/img76/7307/staticsir4.jpg

Homework Equations



i don't know :(

The Attempt at a Solution



I tried but the working is in a notepad, plus I wasnt getting anywhere anyway!

I know its a long questions but this example is essential to my study, any help is largely appreciated!

Thanks in advance!
 
Last edited by a moderator:
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I'll start by teaching you how to do part (i) the center of mass.

1. choose an origin (in this case the best point to choose would be the pivot point)
2. work out the location of the center of mass of the bar and the crate. Both is easy since bar has uniform mass distribution ie. mid-point is the location; and crate can be treated as point mass anyway
3. know your formula:
in x-component of centre of mass
(m_1+m_2)x_{cm}=m_1 x_1 + m_2 x_2
for y:
(m_1+m_2)y_{cm}=m_1 y_1 + m_2 y_2

Masses are given, (x_1,y_1) and (x_2,y_2) are the location of the center of mass of the indivdual m_1 and m_2 which you will have to work out using some geometry (ie. sin, cos)

apart from that you get (as you stated) x_{cm}\approx 1.28m, y_{cm}\approx 0.67m from the pivot point. try it!

I know its a long questions but this example is essential to my study, any help is largely appreciated!

sure it is essential to your study...looks very much like a typical 1st year undergrad exam question
 
Last edited:
Your other questions can be solved with a bit of trig and by using Newton's first law: sum of forces in x direction =0; sum of forces in y direction =0; sum of torques about any point =0. You must show some attempt at your work before we can help further. HINT: Since the wall is frictionless, there can be no vertical force reaction at the wall.
 

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