What is the required force to push a box upwards on an angled surface?

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Homework Help Overview

The problem involves determining the force required to push a box upwards on an angled surface, considering the effects of gravity and friction. The box has a mass of 15 kg and a coefficient of friction of 0.3, with an angle of 35 degrees from the horizontal.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the forces acting on the box, including the normal force and frictional force, and questions whether the box would slip if no additional force is applied. Participants question the calculations related to the downward force and suggest corrections.

Discussion Status

Participants are engaged in clarifying the calculations presented by the original poster. Some guidance has been offered regarding the correct expression for the downward force, indicating a productive direction in the discussion.

Contextual Notes

There is a mention of specific values for mass, friction, and angle, but the discussion also reflects uncertainty regarding the calculations and assumptions made about the forces involved.

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Homework Statement


A box is placed on an angeled surface.

m = 15 kg
f = 0,3

[PLAIN]http://img5.imageshack.us/img5/8717/friction.png

Will the box slip downwards if X = 0?

How big does the force X have to be to push th box upwards?

Homework Equations


The Attempt at a Solution


My idea was that the only force working upwards is Ff.

A = 55°
g = 9,81
N = mg*cos(90-35)+18 = 102,40 N

Ff = 0,3*N = 30,72

Force downwards: mgcos35-18 = 102,54 N

The force down is greater than Ff therefore it would slip if X = 0.To get how big X needs to be to make the box move up i simply subtracted the force down with Ff and got = 71,82 N.Could this possibly be correct?
 
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Why is your downward force mgcos(35)-18? It should just be mgcos(35)
 
Pi-Bond said:
Why is your downward force mgcos(35)-18? It should just be mgcos(35)


Yea, okey. So the rest would be correct then?
 
After changing the value of the downward force, yes.
 

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