Max Velocity of Truck on Banked Track: Spider at Rest

[stateofdogma]

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

F = \frac{mv^2}{r}

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.

 

[haruspex]

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.)

 

[stateofdogma]

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

 

[haruspex]

It isn't stated, but I gather the spider is on the outside of the bend wrt the truck.

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?

 

[stateofdogma]

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

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.

 

[haruspex]

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.

 

[stateofdogma]

alright, I shall move on thanks