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

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To move a 100kg box up a 20-degree ramp, the static and kinetic friction coefficients are Us=0.9 and Uk=0.6, respectively. The maximum pushing force of 1000N is sufficient to overcome the forces required to move the box, calculated as 335N for kinetic friction and 828.8N for static friction. If the box is stopped on the ramp, the force needed to start moving it again is 828.8N, which is still less than the maximum pushing force. Understanding the components of gravitational force, specifically using cosine for the normal force on an incline, is crucial for accurate calculations. Overall, the box can be moved without assistance if a running start is taken, and it can be restarted after stopping.
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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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