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Banked Curve

  • Thread starter Nicolaus
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  • #1
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Homework Statement


How would I go about solving this Newtonian problem?
A truck is going around a circular track of radius 72m, banked at 60degrees. A spider rests on the inside wall of the truck. The coefficient of static friction b/w truck wall and spider is .91. Find the max speed that the truck can have before the spider begins to slip down wall.
I attempted using the reference frame of the spider inside the truck and resolving all forces acting on it, and equating it to the centripetal force of course, but I arrive at the supposedly wrong answer.
The spider is the small circle inside the rectangle in the attachment.


Homework Equations





The Attempt at a Solution

 

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  • #2
haruspex
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I cannot open the .docx, and since it is showing "0 views", I'm guessing no-one else can either. Is the spider on wall of the truck nearest the inside of the bend?
Please post your attempt at solution.
 
  • #3
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The spider is latched onto the tilted surface perpendicular to the inclined ramp (and the driver in his seat in the truck). I resolved, into components, the forces acting on the spider and arrive at an answer of 12.xx m/s. Since acceleration is centripetal (pointing towards the centre), I resolved the forces into a horizontal x and vertical y plane, with respect to the horizontal ground, NOT along the incline.
 
  • #4
haruspex
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The spider is latched onto the tilted surface perpendicular to the inclined ramp (and the driver in his seat in the truck). I resolved, into components, the forces acting on the spider and arrive at an answer of 12.xx m/s. Since acceleration is centripetal (pointing towards the centre), I resolved the forces into a horizontal x and vertical y plane, with respect to the horizontal ground, NOT along the incline.
Yes, but am I right to assume that 'down' for the spider is towards the centre of the turn?
When I asked you to post your attempt, I meant your working. This is a standard requirement on these forums.
 
  • #5
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Down for the spider is parallel to the incline, so 60degrees with respect to the horizontal (in the same direction as mgsin(theta).
This is what I did (forces acting on spider):
Fnetx = N(μs sinθ -cosθ) = mv^2/r
Fnety = 0 = Nsinθ +Ffcosθ - mg --> ∴ N = mg/(sinθ + μcosθ)
substitute for N and rearrange for v.
 
  • #6
haruspex
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You're stil not grasping what it is that's ambiguous in your description. If I am standing in the truck looking at the spider, am I facing the centre of the turn or do I have my back to it? I'm guessing the former.
Based on that, your equations and answer look right.
Are you quite sure it's 60 degrees to the horizontal, not the vertical?
 
  • #7
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Yes, you would be facing the centre of the turn. Sorry for the ambiguity.
Thanks.
 
  • #8
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Is it correct to say that, if the track is banked 60 degrees, then the side wall that the bug is on is oriented 60 degrees to the vertical (30 degrees to the horizontal)? If this were the case, then, in your equations, θ = 60 degrees. Correct? And the bug is on the wall of the truck that is on the inside of the turn?

Chet
 
  • #9
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@Chestermiller: Yes, that's correct.
 
  • #10
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@Chestermiller: Yes, that's correct.
Then I agree with your analysis. How does your final result compare with the "right" answer?
 

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