# Banked Curve

Nicolaus

## 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.

## The Attempt at a Solution

#### Attachments

• PHYS QUESTION.docx
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Homework Helper
Gold Member
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?

Nicolaus
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.

Homework Helper
Gold Member
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.

Nicolaus
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.

Homework Helper
Gold Member
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.
Are you quite sure it's 60 degrees to the horizontal, not the vertical?

Nicolaus
Yes, you would be facing the centre of the turn. Sorry for the ambiguity.
Thanks.

Mentor
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

Nicolaus
@Chestermiller: Yes, that's correct.

Mentor
@Chestermiller: Yes, that's correct.