What is the stretch of the spring holding a wedged board?

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SUMMARY

The problem involves calculating the stretch of a spring holding an 11.5 kg board wedged at a 50.0° angle. Using Hooke's Law, represented by the equation F = -kx, where k is the spring constant of 206 N/m, the force exerted by the board due to gravity can be calculated. The correct approach involves determining the vertical component of the gravitational force acting on the board and then using this force to find the spring's stretch.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Basic knowledge of trigonometry (sine function)
  • Familiarity with gravitational force calculations (weight = mass x gravity)
  • Ability to resolve forces into components
NEXT STEPS
  • Calculate the gravitational force acting on the board using F = mg
  • Learn how to resolve forces into vertical and horizontal components
  • Study the application of Hooke's Law in real-world scenarios
  • Explore problems involving springs and inclined planes for further practice
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and spring dynamics, as well as educators looking for practical examples of force resolution and Hooke's Law applications.

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Homework Statement


A 11.5 kg board is wedged into a corner and held by a spring at a 50.0° angle, as the drawing shows. The spring has a spring constant of 206 N/m and is parallel to the floor. Find the amount by which the spring is stretched from its unstrained length.

Homework Equations



F=-kx

The Attempt at a Solution



I tried doing sin50(11.5kg)/-206 but that didn't work
 
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Can you post the drawing?
 
I tried but it didnt work. I'm not sure how to do it.
 

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