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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.

- when the ramp is frictionless.
- when the coefficient of kinetic friction is 0.18

## 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**

** 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