A curve of radius 164 m is banked at an angle of 12°. An 776-kg car negotiates the curve at 86 km/h without skidding. Neglect the effects of air drag and rolling friction. Find the following.

So far Im just trying to calculate Normal force of pavement on the car tires. Ive calculated it at 7.78kN but the answer says 8.01N my equation for Fn Fn=774kgx9.81/cos9 the text even gives the equation Fn=mg/cos for sample problem.

Question

A curve of radius 164 m is banked at an angle of 12°. An 776-kg car negotiates the curve at 86 km/h without skidding. Neglect the effects of air drag and rolling friction. Find the following.

So far Im just trying to calculate Normal force of pavement on the car tires.  Ive calculated it at 7.78kN but the answer says 8.01N my equation for Fn  Fn=774kgx9.81/cos9  the text even gives the equation Fn=mg/cos for sample problem.

anonymous · Answer

typo i used cos12

anonymous · Answer

Hope  this helps

anonymous · Answer

That is a little confusing, it doesnt explain why the equation is supposedly wrong.

anonymous · Answer

is there friction in this problem?

anonymous · Answer

no. the problem w the equation is the fact that the things in circular motion. You know the upward force is mg and the sideways must be m(v^2)/r. square these numbers, add them and then take the square root and you should get 8.01

anonymous · Answer

What is the normal force then?

anonymous · Answer

mg/Cos(theta) = N is what I have

anonymous · Answer

I have two books that say that is the normal force equation for this particular instance.  Fn= 776x9.81/cos12

anonymous · Answer

glad  you solved it

anonymous · Answer

yeah but that gives us 7.76kN  and the correct answer is 8.01kN.

anonymous · Answer

yes the formula is only for a body at rest though. it doesnt apply.

anonymous · Answer

the upward force of course must be mg. and the sideways force for a thing going around in a circle is given by m(v^2)/r.  these are the legs of the triangle. square them add them and square root the sum to get the correct normal force which is 8.01kN

anonymous · Answer

Fnsin12(776x23.888m/x)^2/164, solve for Fn gives us 13.0kN

anonymous · Answer

alright I will try that and i got 8.07kN

anonymous · Answer

ok. right. thats because the total force isnt just normal but also frictional. take the ration of the centripital force to the vertical force, take the inverse tangent of this to find the angle in which the total force is directed, subtract 12 to find the diference between that and that of the normal force. take the cos of this number and multiply it by 8.07 and i got 8.00

anonymous · Answer

and I assume I was off due to rounding