Box being pushed up an inclined plane

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SUMMARY

The discussion centers on calculating the normal force and coefficient of kinetic friction for a 29 kg box being pushed up a 27-degree inclined plane with a horizontal force of 250 N. The gravitational force acting on the box is determined to be 284.2 N. The normal force, which acts perpendicular to the inclined surface, is calculated to be 370 N, while the coefficient of kinetic friction is found to be 0.26. The correct answer is option e, confirming the calculations and vector resolutions discussed.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of vector resolution in physics
  • Familiarity with the concepts of normal force and friction
  • Basic proficiency in trigonometry, particularly with angles and triangles
NEXT STEPS
  • Study the principles of inclined planes in physics
  • Learn about the derivation and application of the friction equation Fk = μk Fn
  • Explore vector resolution techniques in two-dimensional motion
  • Investigate the effects of different angles of inclination on normal force and friction
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and forces, as well as educators seeking to clarify concepts related to inclined planes and friction.

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Homework Statement



A 250N force is directed horizontally to push a 29kg box up a plane that is inclined at 27 degrees to the horizontal. the box moves up the plane at a constant speed. Determine the magnitude of the normal force, and the coefficient of kinetic friction.
g=9.8m/s^2
a(y) = 0m/s

Homework Equations


F=mg to find the force that is pulling down on it
Fk= \mu k Fn


The Attempt at a Solution



I understand that the forces need to be resolved into the same direction, but I'm not sure what direction the normal force would be pointing in. The force of gravity i got was 284.2N. I can't figure out the kinetic friction until i figure out the normal force.
 
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The normal force will be at 90 degrees to the inclined surface, and equal and opposite to the component of the weight of the box acting perpendicularly through the plane.
 
maybe I'm resolving my vectors wrong. when i resolve the vector for weight, the part that should be equal and opposite to Fn I'm getting 129.02N which is not possible as it is multiple choice and that's not even one of the answers
 
What are the options you are given?
 
a) 330N 0.31
b) 310N 0.33
c) 250N 0.27
d) 290N 0.30
e) 370N 0.26

The answer is e
 
To find the components of the weight, draw an angle formed by the vector of weight and a line perpendicular to and under the surface. This angle is equal to the angle of inclination of the plane.
Make the x-axis parallel to the surface, so the y-axis would be perpendicular to the surface and parallel to the normal force.
 

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