A box with the mass m = 25 kg is sliding up a hill

  • #1
Danielheidarr
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Homework Statement
A box with the mass m = 25 kg is sliding with starting speed of V0=7m/s up a hill that has the angle Θ=12°. there is a steady wind blowing with a horizontal force of Fv = 50N. The coefficient of friction is µk = 0.18.
Relevant Equations
a) What is the force that works on the box from the hill?
b) What distance, L, does the box slide up the hill before it stops?
c) What does the coefficient of friction need to be at the lowest so the box doesn´t slide back down the hill*
What I have already attempted is on my Ipad and I don't know how to upload it on here.
With hope of help
DJ
 
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  • #2
Use my tips from your other thread to use "Attach files". :wink:
 
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  • #3
Danielheidarr said:
there is a steady wind blowing with a horizontal force of Fv = 50N.
Which way?
 
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  • #4
haruspex said:
Which way?
Upp the hill (with the box)
 
  • #5
Danielheidarr said:
Upp the hill (with the box)
You mean to say that there is an uphill component. The 50 N wind is horizontal.
 
  • #6
If the wind is blowing down a slope, I guess it can't be horizontal.

Blowing horizontally into the slope is perhaps more precise than up the slope.
 
  • #7
kuruman said:
You mean to say that there is an uphill component. The 50 N wind is horizontal.
yeah that's what I meant sorry
 
  • #8
kuruman said:
You mean to say that there is an uphill component. The 50 N wind is horizontal.
sorry the uphill component is only for the box, the box is moving to the right towards the hill and the wind is moving horizontal in the same direction as the box.
 
  • #9
PeroK said:
If the wind is blowing down a slope, I guess it can't be horizontal.

Blowing horizontally into the slope is perhaps more precise than up the slope.
yeah what i meant to say is blowing horizontally into the slope.
 
  • #10
Note 17 Nov 2022.jpg
 
  • #11
This is what I did. What do you guys think?
 
  • #12
Here is a pdf probably better to view that
 

Attachments

  • Note 17 Nov 2022.pdf
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  • #13
Danielheidarr said:
This is what I did. What do you guys think?
##6.5m## looks like the distance without the wind blowing.
 
  • #14
ah that's right i forgot to add that in thank you so much :)
 
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  • #15
Same for part c). ##\mu = \tan \theta ## is the calculation without the wind.
 
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  • #16
Same for part a.
 
  • #17
PeroK said:
Same for part a.
I'm not quite sure how to work the wind into the equation. Any hints for it?
 
  • #18
Danielheidarr said:
I'm not quite sure how to work the wind into the equation. Any hints for it?
It's a force like any other.
 
  • #19
Danielheidarr said:
I'm not quite sure how to work the wind into the equation. Any hints for it?
See post #5.
 
  • #20
Danielheidarr said:
This is too hard to read. The subscripts are illegible and the thumbnail sketch FBD far too small. Draw decently sized diagrams and type equations.
Don't forget the wind force has a component normal to the slope.
 
  • #21
haruspex said:
This is too hard to read. The subscripts are illegible and the thumbnail sketch FBD far too small. Draw decently sized diagrams and type equations.
Don't forget the wind force has a component normal to the slope.
I also sent the pdf in post #12 better view there
 
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