Solving Horizontal Tension: Step-by-Step Guide

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SUMMARY

The discussion focuses on solving the equation TLsinO - mg(1/2LcosO) = 0, which relates to torque calculations in physics. Participants clarify that torque (t) is calculated using the formula t = fd, where d represents the perpendicular distance from the pivot point to the line of action of the force. The expressions TLsinO and mg(1/2LcosO) represent the torques exerted by tension and weight, respectively, about a hinge. Understanding these concepts is crucial for accurately analyzing forces in angled beams.

PREREQUISITES
  • Understanding of torque and its calculation
  • Familiarity with trigonometric functions in physics
  • Knowledge of forces acting on beams
  • Basic principles of static equilibrium
NEXT STEPS
  • Study the principles of static equilibrium in mechanics
  • Learn about torque calculations in various contexts
  • Explore trigonometric functions and their applications in physics
  • Examine real-world examples of forces acting on beams and structures
USEFUL FOR

Students in physics, engineers working with structural analysis, and anyone interested in understanding torque and forces in mechanical systems will benefit from this discussion.

fldk31
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Hi guys, I came across this question and I was wondering if someone could explain how we get:
TLsinO - mg(1/2LcosO) = 0
Your help would be much appreciated.

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Do you know how to calculate a torque?
 
t=fd and t-wa ?
 
What does d represent there?
 
distance...
 
Not just any distance. It's specifixally the perpendicular distance from the wall. Since the beam is at an angle, you need the trig equations to find the perpendicular distance.
 
If you understand this, what's the problem to explain it in a couple sentences instead of playing games? If you don't want to explain, it's totally fine.
I am so confused what these two parts mean:
1) TLsinO
2) mg(1/2LcosO)
 
fldk31 said:
If you understand this, what's the problem to explain it in a couple sentences instead of playing games? If you don't want to explain, it's totally fine.
I am so confused what these two parts mean:
1) TLsinO
2) mg(1/2LcosO)
If you are familiar with torque then you should know this: if a force F acts through a point P, the torque the force has about a point Q is F x distance PQ x sin(angle between PQ and the direction of the force).
Can you relate that to the diagram and see how the expressions you quote correspond to the torque exerted by T about the hinge, and the torque the weight of the arm exerts about the hinge?
 

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