Scale Reading of Normal Force: Due by Midnight

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SUMMARY

The discussion centers on calculating the normal force reading on a scale when a 61 kg student stands on it while rolling down a 29° incline. The normal force is determined by analyzing the free body diagram (FBD) and breaking down the gravitational force into components. The acceleration due to gravity is specified as 9.81 m/s². The key takeaway is that the normal force reading on the scale will not equal the student's weight due to the incline's angle.

PREREQUISITES
  • Understanding of free body diagrams (FBD)
  • Knowledge of gravitational force and its components
  • Familiarity with Newton's laws of motion
  • Basic trigonometry for resolving forces
NEXT STEPS
  • Learn how to resolve forces in inclined planes using trigonometric functions
  • Study the concept of normal force in different scenarios, including frictionless surfaces
  • Explore advanced applications of free body diagrams in dynamic systems
  • Investigate the effects of varying angles on normal force calculations
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone interested in understanding forces on inclined planes and their applications in real-world scenarios.

IBdoomed
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Remember that a scale records the value of
the normal force, not a person’s actual weight.
Draw a FBD. Rotate your coordinate system.

A 61 kg student weighs himself by standing
on a scale mounted on a skateboard that is
rolling down an incline, as shown. Assume
there is no friction so that the force exerted
by the incline on the skateboard is normal to
the incline.
The acceleration of gravity is 9.81 m/s

What is the reading on the scale if the angle
of the slope is 29◦?
Answer in units of N

I have my free body diagram with Fn going up and to the left and Fg gong straight down. I do not understand how to rotate the FBD or what that would accomplish...
ANY HELP WOULD BE MUCH APPRECIATED!
 
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Your FBD is correct. Now break up the weight force into components parallel and perpendicular to the plane, and cross out the initial weight vector. The sum of forces perpendicular to the incline must equal zero, since the skateboarder does not lift off the incline in that direction.
 
much thanks! mission accomplished
 

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