Calculate Weight Component on Inclined Plane

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SUMMARY

The discussion focuses on calculating the weight component of a box on an inclined plane at a 30-degree angle. The correct formula to determine the parallel component of weight is W*sin(θ), where W represents the weight of the box. Given that sin(30) equals 0.5, the magnitude of the weight component acting parallel to the incline is 0.5W. This conclusion simplifies the problem significantly, confirming that the answer is indeed 0.5W.

PREREQUISITES
  • Understanding of basic physics concepts, specifically forces and weight.
  • Familiarity with trigonometric functions, particularly sine.
  • Knowledge of vector representation of forces.
  • Ability to manipulate equations involving weight and gravitational force.
NEXT STEPS
  • Study the derivation of weight components on inclined planes using different angles.
  • Learn about the implications of friction on inclined planes in physics.
  • Explore the application of Newton's laws in inclined plane problems.
  • Investigate the use of vector decomposition in solving physics problems.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined plane problems, as well as educators looking for clear examples of vector decomposition in real-world applications.

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Homework Statement


the weight of a box on a plane inclined at 30 def is represented by the vector W. What is the magnitude of the component of the weight that acts parallel to the incline.
Answer is given in terms of W
possible anwsers .5W 1.5W .87W and W


Homework Equations





The Attempt at a Solution


M x g x sin(30)= 4.9
??
not sure what formula is needed to be used here
 
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W = mg. So
W*sinθ = ?
 
You don't know M, but you do know that W = M*g
so you have W*sin(30) = 0.5*W
 
wow i made it so much harder than what it needed to be thanks
 

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