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

AI Thread Summary
The discussion revolves around a box with a mass of 25 kg sliding up a hill while experiencing a horizontal wind force of 50 N. Participants clarify that the wind's horizontal force contributes an uphill component for the box as it moves towards the hill. There are requests for assistance in incorporating the wind force into the calculations, with reminders that it acts like any other force and has a normal component to the slope. Concerns are raised about the clarity of diagrams and equations shared, suggesting improvements for better readability. Overall, the conversation focuses on understanding the dynamics of the box's movement in relation to the wind force.
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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Use my tips from your other thread to use "Attach files". :wink:
 
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Danielheidarr said:
there is a steady wind blowing with a horizontal force of Fv = 50N.
Which way?
 
haruspex said:
Which way?
Upp the hill (with the box)
 
Danielheidarr said:
Upp the hill (with the box)
You mean to say that there is an uphill component. The 50 N wind is horizontal.
 
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.
 
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
 
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.
 
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
 

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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 :)
 
  • #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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