Equations of Equilibrium - First year engineering

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SUMMARY

The discussion focuses on calculating the reaction forces FA and FB for a uniform beam supported by a roller at point A and a pin at point B, subjected to external forces of 75.0N and 58.0N. The beam has a mass of 24.0kg, resulting in a weight of 235.2N. The calculations yield FA = 270N and FB = 290N, derived from equilibrium equations: ∑FX = 0, ∑FY = 0, and ∑MO = 0. Participants emphasize the importance of correctly applying trigonometric functions to resolve forces at angles.

PREREQUISITES
  • Understanding of static equilibrium principles in mechanics
  • Proficiency in using trigonometric functions for force resolution
  • Familiarity with Newton's laws of motion
  • Basic knowledge of beam mechanics and reaction forces
NEXT STEPS
  • Study the application of static equilibrium in beam analysis
  • Learn how to resolve forces using trigonometric identities
  • Explore the concept of moment arms in calculating torques
  • Practice problems involving reaction forces in various support configurations
USEFUL FOR

First-year engineering students, mechanics instructors, and anyone studying static equilibrium and beam analysis in structural engineering.

aidan.s
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Homework Statement



As shown, a roller at point A and a pin at point B support a uniform beam that has a mass 24.0kg . The beam is subjected to the forces F2 = 75.0N and F2 = 58.0N . The dimensions are L1 = 0.550m and L2 = 1.70m . What are the magnitudes FA and FB of the reaction forces FA and FB at points A and B, respectively? The beam's height and width are negligible.picture:
http://session.masteringengineering.com/problemAsset/1180309/3/1118668_004.jpg"

Homework Equations



[tex]\sum[/tex]FX = 0
[tex]\sum[/tex]FY = 0
[tex]\sum[/tex]MO = 0

The Attempt at a Solution



24.0 kg * 9.8 = 235.2N of force at (.55+1.7)/2 = 1.125m

trying to solve for the normal force at A, NA **anticlockwise = positive**
(75N * 1.7m) + (235.2N * 1.125m) - (2.25m * NA * 3/5) = 0
NA = 270N

solving for BX
sum of forces in x direction = 0; **positive direction to the right**
-BX - 58.sin(15) + (NA*4/5) = 0
BX = 200

solving for BY
sum of forces in the y direction = 0l **positive direction is upwards**
-235.2 - 75 - 58.cos(15) + (NA * 3/5) + BY = 0
BY = 204

so as the question asks, FA = 270N ? is this correct and;
FB = [tex]\sqrt{}[/tex] 200^2 + 204^2 = 290N ? - is this the right way to find the magnitudes of the force?this is the basic method I've been trying with a few minor variants but this program says there all wrong so I'm completely lost, I'm not expecting anyone to give me a straight answers but some direction as to how I've gone wrong would be great

im pretty sure its just the fact that the roller at point A is on an angle is what's stuffing me up

thanks, aidan
 
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You may have your 3/5 and your 4/5 swapped (3 places) - check.
 
./facedesk

thanks, i think I am going to stick to just converting them triangles into actual angles and use sin n cos otherwise i keep stuffing them up :/

cheers though yeah :)