Box Pushed Up an Incline Ramp: Basic Newton's Law Questions

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Homework Help Overview

The discussion revolves around a problem involving a box being pushed up an inclined ramp using Newton's laws. The scenario includes analyzing forces acting on the box, particularly in a frictionless context and with kinetic friction present.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of forces acting on the box, including gravitational force and normal force. There is uncertainty regarding the correct application of these forces and whether the approach to summing forces is accurate.

Discussion Status

Some participants have provided feedback on the drawing of the normal force, suggesting that it may not have been represented correctly. This has led to further clarification about the relationship between the normal force and the components of gravity acting on the box.

Contextual Notes

There is mention of a coefficient of kinetic friction and the need to consider both frictionless and friction scenarios. Participants are also reflecting on the implications of their force diagrams and calculations.

Kathy W
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Homework Statement


A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force.
  1. when the ramp is frictionless.
  2. when the coefficient of kinetic friction is 0.18
upload_2015-7-22_11-43-55.png

Homework Equations


F=mg
FN=mg
μ= Fk/FN

The Attempt at a Solution


a) mg=(90)((.8)= 882N=Fg

sin28=Fgx/882N
(0.469)(882)= Fgx
414N=Fgx= magnitude of applied force

b) FN=mg= 882N
μ=Fk/FN
0.18=Fk/882
(882)(0.18)= Fk
Fk= 158.76

** This is where I am unsure of my approach**

Fapplied= Fgx+Fk
=414N+158.76 N
= 572N

That seems like too much to me, but I'm not quite sure. Does it look like I have that equation correct ,or am I supposed to subtract the value I found for Fk from FgX, or something totally different. Any help would be very much appreciated
 
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Kathy W said:
FN=mg= 882N
μ=Fk/FN
0.18=Fk/882
(882)(0.18)= Fk
Fk= 158.76

You have not drawn FN correctly . Redraw correctly and try again .

Hope this helps .
 
Qwertywerty said:
You have not drawn FN correctly . Redraw correctly and try again .

Hope this helps .
oh okay, is it opposing the Fy aspect instead of Fg? I was confused when I was drawing that! Thank you for your response
 
Kathy W said:
oh okay, is it opposing the Fy aspect instead of Fg? I was confused when I was drawing that! Thank you for your response
Yes. The normal force opposes that component of gravity which would cause acceleration perpendicular to the surface ( the incline plane).
 

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