Tension in string at a point on a frame

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SUMMARY

The discussion focuses on calculating tension in a frame with a pulley system using basic trigonometric principles. The user correctly applies the equations F = mg and Tx = mg*sin(theta) to determine the tension, resulting in a value of 637N for the total tension and 450.4N for the components Tx and Ty at a 45-degree angle. The solution is confirmed as accurate, indicating that the problem can be solved with straightforward calculations without additional complexities related to frame support.

PREREQUISITES
  • Understanding of basic physics principles, specifically Newton's laws of motion.
  • Familiarity with trigonometric functions and their application in physics.
  • Knowledge of tension calculations in static systems.
  • Ability to perform unit conversions and apply them in problem-solving.
NEXT STEPS
  • Explore advanced tension analysis in dynamic systems involving pulleys.
  • Learn about the implications of different angles on tension in various configurations.
  • Study the effects of friction in pulley systems and how it alters tension calculations.
  • Investigate the use of vector components in more complex mechanical systems.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and professionals involved in engineering and design of mechanical systems involving tension and pulleys.

LeafMuncher
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Homework Statement


btMUsJA.jpg

Hi all. What concerns me here is that it's worth 10 marks, but the solution I've tried only takes 2 steps. Am I missing some relation between the frame support and the pulley altering the components of the tension, or is the solution really just a basic trig conversion?

Homework Equations


F = mg
Tx = mg*sin(theta)

The Attempt at a Solution


using the above conversions I get the tension as 65kg*9,8m/s^2 = 637N
Then just convert using the 1/1 ratio as 45deg, giving Tx and Ty = 637*sin(45) = 450.4N
 
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LeafMuncher said:

Homework Statement


btMUsJA.jpg

Hi all. What concerns me here is that it's worth 10 marks, but the solution I've tried only takes 2 steps. Am I missing some relation between the frame support and the pulley altering the components of the tension, or is the solution really just a basic trig conversion?

Homework Equations


F = mg
Tx = mg*sin(theta)

The Attempt at a Solution


using the above conversions I get the tension as 65kg*9,8m/s^2 = 637N
Then just convert using the 1/1 ratio as 45deg, giving Tx and Ty = 637*sin(45) = 450.4N
Your solution is correct. Good job.
 

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