Max Velocity of Truck on Banked Track: Spider at Rest

In summary, the conversation discusses the maximum velocity of a truck on a circular track with a spider at rest on its side. The problem involves using the equations for centripetal force and friction to determine the velocity at which the spider will remain at rest. The solution involves breaking down the forces on different axes and solving for the velocity using the radius and gravitational force.
  • #1
stateofdogma
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Homework Statement


A Truck is going around a circular track of radius 72m and is banked at 60 deg. A spider is at rest on the side of the truck and the coeffiecient of friction is 0.91, what is the maximum velocity of the truck so that the spider is at rest.


Homework Equations



[tex]F = \frac{mv^2}{r}[/tex]

The Attempt at a Solution


What I did is take the coordinate axes along the wall of the spider, then made -mgsin(%)+ N = -mv^2/2cos(%) for the vertical axes (where (%)is 60 deg ). -mgcos%+ musN = mv^2/2sin% for the horizontal axis. then substitute in N for the horizontal equation and solve for v. (The solution was (mv^2/r)sin60 - mus((mv^2/r)cos60 + mgsin60) = mgcos60, v = 47 m/s When I look at the solution I see that for the vertical axis the centripetal force and gravitational force have the same sign when it equal to Normal force which implies that when you subtract it to from both sides, the normal force is in the same direction as the centripetal force. Isn't the centripetal force just another way of writing the resultant force and should therefore be a result of other forces, instead of contributing itself.
 
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  • #2
stateofdogma said:
What I did is take the coordinate axes along the wall of the spider, then made -mgsin(%)+ N = -mv^2/2cos(%) for the vertical axes (where (%)is 60 deg ).
Please explain how you get the right hand side of that equation.
(I would recommend using vertical and horizontal axes instead, since the centripetal force is horizontal.)
 
  • #3
if you mean -mv^2/rcos60 I reason after breaking the centripetal force on the wall of the truck the direction you get is opposite to the normal force of the spider, making it in the same directin of the component gravitational force
 
  • #4
It isn't stated, but I gather the spider is on the outside of the bend wrt the truck.
stateofdogma said:
if you mean -mv^2/rcos60
Oh, that's ok (presuming you mean (mv^2/r)cos60), but you wrote mv^2/2cos60.

for the vertical axis the centripetal force and gravitational force have the same sign when it equal to Normal force which implies that when you subtract it to from both sides, the normal force is in the same direction as the centripetal force.
Sorry, I cannot follow what you are saying there. In the vertical direction, the centripetal force has no component. In the horizontal, the gravitational has no component.
To get them with the same sign, you must be referring to the direction parallel to the road surface. But then I don't know what you mean by "it equal to the normal force". What is?
 
  • #5
haruspex said:
It isn't stated, but I gather the spider is on the outside of the bend wrt the truck.

Actually from the image the spider is on the wall closer to the ground, that is to say it doesn't have the wall of the truck pushing on it in the centripetal motion. I have uploaded an image


haruspex said:
Sorry, I cannot follow what you are saying there. In the vertical direction, the centripetal force has no component. In the horizontal, the gravitational has no component.
To get them with the same sign, you must be referring to the direction parallel to the road surface. But then I don't know what you mean by "it equal to the normal force". What is?

Thats because I broke up the radial acceleration on the wall, but I'll use the axis along the radial acceleration. here what I get

fsin(60) - Ncos(60) = (mv^2/r) for the horizontal axis f for friction force

fcos(60) + Nsin(60) = mg for the vertical axis

then I get v = sqrt( (musin(60) - cos(60)/( mucos(60) + sin(60)))*Rg R for radius

V= 12.4 m/s , but the answer is 40.7 m/s, I upload how the solution was done.
 

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  • #6
stateofdogma said:
Actually from the image the spider is on the wall closer to the ground, that is to say it doesn't have the wall of the truck pushing on it in the centripetal motion. I have uploaded an image
OK, I was thinking it was on the outside of the truck, outside of the bend, but inside-inside is the same.
Thats because I broke up the radial acceleration on the wall, but I'll use the axis along the radial acceleration. here what I get

fsin(60) - Ncos(60) = (mv^2/r) for the horizontal axis f for friction force

fcos(60) + Nsin(60) = mg for the vertical axis

then I get v = sqrt( (musin(60) - cos(60)/( mucos(60) + sin(60)))*Rg R for radius

V= 12.4 m/s , but the answer is 40.7 m/s, I upload how the solution was done.
I agree with your answer. The book answer seems to have some signs wrong (the cos 60 and sin 30 terms) Intuitively, the spider wasn't that far off slipping in a stationary truck, so it is not going to take much speed to dislodge it.
 
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  • #7
alright, I shall move on thanks
 

1. What is the purpose of determining the maximum velocity of a truck on a banked track?

The purpose of determining the maximum velocity of a truck on a banked track is to understand the limits of the truck's speed while traveling on a curved surface. This information can be used to ensure safe driving practices and to optimize the design of banked tracks for different vehicles.

2. How is the maximum velocity of a truck on a banked track calculated?

The maximum velocity of a truck on a banked track is calculated using the equation v = √(rg tanθ), where v is the velocity, r is the radius of the track, g is the acceleration due to gravity, and θ is the angle of the banked track.

3. What factors affect the maximum velocity of a truck on a banked track?

The maximum velocity of a truck on a banked track is affected by the angle of the banked track, the radius of the track, the weight and size of the truck, and the coefficient of friction between the tires and the track surface.

4. Can a truck achieve the maximum velocity on a banked track if it is not at rest?

No, a truck cannot achieve the maximum velocity on a banked track if it is not at rest. The maximum velocity is only achieved when the truck is at rest on the banked track and has enough centripetal force to maintain its speed while traveling on the curved surface.

5. How does the maximum velocity of a truck on a banked track compare to a truck on a flat track?

The maximum velocity of a truck on a banked track is higher than that on a flat track. This is because the angle of the banked track helps to provide additional centripetal force, allowing the truck to maintain a higher speed without slipping or skidding.

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