Friction Force on Banked Curves

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Homework Statement




A car weighing 3220 lbs rounds a curve a 200 ft radius banked at an angle of 30deg. Find the friction force acting on the tires when the car is traveling at 60mph. Coefficient of friction is 0.9.



Homework Equations




i rotated the axes such that y-axis is collinear with Fn

and x-axis is collinear with Ff


I tried equilibrating the y-components i get Fn = 4,724.6 lb , then Ff=(0.9)(4,724.6) = 4252.14 lb


while on equilibrating the x-components i got Ff= 1743.25lb



what does this implicate? is that possible? Please help. big Thanks!

PS. I have read from my book that Ff=(u)(Fn) is only true if skidding is impending or v is at maximum.. how come? so what's the meaning og 4252.14?



The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Doc Al
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i rotated the axes such that y-axis is collinear with Fn

and x-axis is collinear with Ff
Why do that? It might not be the best idea here, since there is a horizontal acceleration.
I tried equilibrating the y-components i get Fn = 4,724.6 lb , then Ff=(0.9)(4,724.6) = 4252.14 lb


while on equilibrating the x-components i got Ff= 1743.25lb
Instead, identify the forces in a FBD. Then apply Newton's 2nd law to the horizontal and vertical components.
PS. I have read from my book that Ff=(u)(Fn) is only true if skidding is impending or v is at maximum.. how come?
That's correct. For static friction, μFn gives the maximum possible friction force. So you cannot use that here, since there's no reason to think that you are at the limit of static friction. Instead, call that friction force Ff and solve for it.
 
  • #3
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Thanks for reply sir! I still some blurry ideas in my mind but i wanna gather it first to have a better follow question :)
 
  • #4
Doc Al
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Thanks for reply sir! I still some blurry ideas in my mind but i wanna gather it first to have a better follow question :)
Good. Get to work! :smile:
 
  • #5
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Another question sir! What if i got that Fn= 4724 lb then mutiplied it by u=.9 . What does that quantity mean? does it mean something at all? or it does not exist?
 
  • #6
Doc Al
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Another question sir! What if i got that Fn= 4724 lb then mutiplied it by u=.9 . What does that quantity mean? does it mean something at all? or it does not exist?
First a question for you: How did you arrive at Fn = 4724 lb?
 
  • #7
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I used FBD. I get Ff= 1743 and Fn= 4724.
 
  • #8
Doc Al
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I used FBD. I get Ff= 1743 and Fn= 4724.
Show me exactly what you did.
 
  • #9
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ahh..haha, i'm sorry for that silly answer.haha.. here's my solution

Summation of forces horizontal

(Fn)Sin30 + (Ff)(Cos30) = Fc
(Fn)(sin30) + (ff)(cos 30)= (Wv^2)/(gr)
(Fn)(sin30) + (ff)(cos 30)= ((3220)(88^2))/((32.2)(200))

(Fn)(sin30) + (ff)(cos 30)= 3872 <--------Equation (1)



Summation of Forces vertical

(Fn)(Cos30) = (Ff)(Sin30) + W

(Fn)(cos30) - (Ff)(sin30) = 3220 <--------Equation (2)



Solving this two equations simultaneously we get, Fn= 4724 and Ff = 1743


From there, If i used the fact that Ff=(u)(Fn) could not be used, I conclude that Ff=1743 must be the true value of the Frictional Force not Ff=(0.9)(4724) since it does not impend to skid nor have the maximum velocity.

What bothers me is that value, the Ff=(0.9)(4724), what does it mean? does it really mean something ? does that frictional force exist? or what? :)
 
  • #10
Doc Al
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ahh..haha, i'm sorry for that silly answer.haha.. here's my solution

Summation of forces horizontal

(Fn)Sin30 + (Ff)(Cos30) = Fc
(Fn)(sin30) + (ff)(cos 30)= (Wv^2)/(gr)
(Fn)(sin30) + (ff)(cos 30)= ((3220)(88^2))/((32.2)(200))

(Fn)(sin30) + (ff)(cos 30)= 3872 <--------Equation (1)



Summation of Forces vertical

(Fn)(Cos30) = (Ff)(Sin30) + W

(Fn)(cos30) - (Ff)(sin30) = 3220 <--------Equation (2)



Solving this two equations simultaneously we get, Fn= 4724 and Ff = 1743
I didn't check your arithmetic, but that's exactly the right approach.

(So you didn't really 'rotate the axes', despite what you said in your first post. Good.)
From there, If i used the fact that Ff=(u)(Fn) could not be used, I conclude that Ff=1743 must be the true value of the Frictional Force not Ff=(0.9)(4724) since it does not impend to skid nor have the maximum velocity.
Good.

What bothers me is that value, the Ff=(0.9)(4724), what does it mean? does it really mean something ? does that frictional force exist? or what? :)
μN represents the maximum possible static friction between two surfaces with a given normal force. It's not particularly useful in this problem.

An example where it might be useful would be if you had to figure out the maximum speed that you could safely negotiate a given banked curve. In that case, to find the maximum speed, you would set the friction force to its maximum possible value.
 
  • #11
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(I don't know how to quote , ahaha)

Thanks. Actually I had 2 solutions , one is rotating the axes (because my professor told us that in that sense we can eleminate the use of simultaneous equation) and i also use the unrotated one which fairly the same. nevertheless i got the answer. hehe


From ur last statement, i think i cant use that Ff=(0.9)(4724) for getting the max speed since on getting that Fn=4724 i already used the 60 mph speed. What I did to get the max speed is substituting the Fn equation (from summation of forces in y (i this one i rotated the axis for convinience)) to the (u)(Fn) of the another equation (the summation of forces on x axis). :)
 
  • #12
Doc Al
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Thanks. Actually I had 2 solutions , one is rotating the axes (because my professor told us that in that sense we can eleminate the use of simultaneous equation) and i also use the unrotated one which fairly the same. nevertheless i got the answer. hehe
Actually, rotating the axes is a good idea--as long as you do it properly, which you obviously did. (Most folks forget to take components of the acceleration and just assume that the 'vertical' forces cancel.)

From ur last statement, i think i cant use that Ff=(0.9)(4724) for getting the max speed since on getting that Fn=4724 i already used the 60 mph speed.
Right. You can't use that, since Fn depends on the speed.

What I did to get the max speed is substituting the Fn equation (from summation of forces in y (i this one i rotated the axis for convinience)) to the (u)(Fn) of the another equation (the summation of forces on x axis).
Sounds good to me!
 

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