What is the Normal Force on Tires in a Banked Curve with Friction?

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The discussion focuses on calculating the normal force on tires during a banked curve with friction, given specific parameters: a radius of 150m, a 10-degree bank angle, a car mass of 800kg, and a speed of 85 km/h. Participants highlight that the basic formula Fn = mg/cos10 is insufficient due to the additional horizontal forces from angular motion. The complexity arises from the need to account for both gravitational and frictional forces acting on the tires. Ultimately, the correct approach involves analyzing the forces in both vertical and horizontal components to determine the normal force accurately. The conversation concludes with a resolution on understanding the forces at play.
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Homework Statement



Radius 150m, angle is 10 degree, mass of the car 800kg, speed 85 km/h.

Find the normal force on the tires exerted by the pavement. ( There is friction)



Homework Equations



\Sigma F= m* v^2/r

The Attempt at a Solution



The friction makes it's so complicated. I know that Fn= mg/cos10 will not work.

Can anyone help me ?
 
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nns91 said:

Homework Statement



Radius 150m, angle is 10 degree, mass of the car 800kg, speed 85 km/h.

Find the normal force on the tires exerted by the pavement. ( There is friction)



Homework Equations



\Sigma F= m* v^2/r

The Attempt at a Solution



The friction makes it's so complicated. I know that Fn= mg/cos10 will not work.

Can anyone help me ?

It's not that m*g*Cosθ doesn't work, it's just that there is a horizontal force from the angular motion that is acting as well.

If there is a horizontal force, what is its component that is also normal to the incline?
 
Thanks. I got it.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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