Solving for Moving a 100kg Box on a Ramp | Static and Kinetic Friction

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Homework Help Overview

The discussion revolves around calculating the forces involved in moving a 100kg box up a 20-degree ramp, considering static and kinetic friction. Participants explore the implications of the coefficients of friction provided and the maximum pushing force available.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to calculate the kinetic and static friction forces, questioning the correct application of the normal force on an incline. Some express confusion about the use of sine and cosine in their calculations.

Discussion Status

Several participants have provided calculations for the forces involved, with some clarifying the components of gravitational force acting on the box. There is an ongoing exploration of the correct approach to resolving forces, particularly in relation to the normal force.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the depth of explanations and the use of external resources. Some express uncertainty about visualizing the problem through diagrams.

skittlez411
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force! help me please =)

Homework Statement



It's moving day and you need to push a 100kg box up a 20 degree ramp into the truck. The coefficients of friction for the box on the ramp are Us=0.9 and Uk=0.6. Your largest pushing force is 1000N. Can you get the box into the truck without assistance if you get a running start at the ramp? If you stop on the ramp, will you be able to get the box moving again?

Homework Equations



Fk=Ukmg
F=mgsin20

The Attempt at a Solution



a) Fk=(0.6)(100)(9.8)
=588N

F=mgsin20
=335N

yes, my maximum pushing force is 1000N. 335N<1000N

b) Fs=Usmg
=882N

Yes. I will be able to get the box moving again because I have enough force to overcome the static force.

thank you!o:)
 
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skittlez411 said:
Fk=Ukmg
Careful here! Fk = Uk N, where N is the normal force. Since the mass is on an incline, the normal force is not just mg.
 
ok... so i got

a) Fk=ukmgsin20
= 201.11 N

so if i get a running start i would be able to move the box up w/o assistance.

b) Fs= Usmgsin20
= 301.66 N

If i stop at the ramp, i will be able to get the box moving agian.. 301.66N<1000N

is that it? thanks!
 
N is mgcos 20:smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile:
 
pcdagr8 said:
N is mgcos 20:smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile:

can i ask why? i thought it's sine...:blushing:
 
N is normal reaction ,component of gravity perpendicular to the surface
when you are taking the component of gravity parallel to the surface as mgsin20
how are you missing out that perpendicular component will be mg cos20
just draw a free body diagram and resolve the forces...
 
hmmm... i still don't quite get it. i know how sine and cosine are. but i don't quite get the diagram... sorry!
 
well...i can't help u with this.i don't have any software to make a diagram for you...may be someone else can help...
 
  • #10
ok thanks austronuc and pcdagr8! i get that part now!

a) Fk=Ukmgcos20
= 552.5N

b) Fs=Usmgcos20
=828.8N

so this is what i have right now.
 
  • #11
good job:wink:
 
  • #12
yay! i got it? thanks!
 

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