Question

A highway curve of radius 60m is banked at an 18° angle. What is the maximum speed with which a 1500 kg rubber tired car can take this curve without sliding? Assume the coefficient of static friction between the tires and the road is 1.0. Hint: you must consider carefully which way friction points, up the incline or down the incline. Also, remember that for uniform circular motion problems you always want to use radial, tangential, and z axes.

Answer 1

Answer:

$v=34m/s$.

Explanation:

To solve this problem we need to apply Newton's Second Law to the uniform circular motion, this means, we are going to take in account the radial forces. The forces we have in the x axiss are:

• the horizontal component of the friction $f_{x}=\mu Ncos(\theta)$,
• the horizontal component of normal force $N_{x}=Nsin(\theta)$,

(we do not consider the weight here since it is only exerted in the y axiss) so

$\sum F_{x}=ma_{c}$,

$N_{x}+f_{x}=ma_{c}$ the friction points toward the center of the curve, this is why the car will not slide,

$Nsin(\theta)+\mu Ncos(\theta)=ma_{c}$,

now remembering that

$a_{c}=\dfrac{v^{2}}{r}$

we get

$Nsin(\theta)+\mu Ncos(\theta)=\dfrac{mv^{2}}{r}$,

$N(sin(\theta)+\mu cos(\theta))=\dfrac{mv^{2}}{r}$ (this will be eq 1).

Now, for the y axiss we have:

• vertical component of the friction $f_{y}=\mu Nsin(\theta)$,
• vetical component of the normal force $N_{y}=Ncos(\theta)$,
• weight $mg$.

Where the weight and the vertical component of the friction point downwards.

So

$\sum F_{y}=0$,

$N_{y}-f_{y}-mg=0$,

$Ncos(\theta)-\mu Nsin(\theta)=mg$,

$N(cos(\theta)-\mu sin(\theta))=mg$ (this will be eq 2).

Now we do eq 1 over eq 2:

$\frac{N(sin(\theta)+\mu cos(\theta))}{N(cos(\theta)-\mu sin(\theta))}=\dfrac{mv^{2}}{rmg}$,

simplifiying and solving for $v$

$\frac{(sin(\theta)+\mu cos(\theta))}{(cos(\theta)-\mu sin(\theta))}=\dfrac{v^{2}}{rg}$,

$\frac{rg(sin(\theta)+\mu cos(\theta))}{(cos(\theta)-\mu sin(\theta))}=v^{2}$,

$v=\sqrt{\frac{rg(sin(\theta)+\mu cos(\theta))}{(cos(\theta)-\mu sin(\theta))}}$,

$v=\sqrt{\frac{60*9.8(sin(18)+1*cos(18))}{(cos(18)-1*sin(18))}}$,

and finaly

$v= 34m/s$

Answer 2

Answer:

The speed is 33.98 m/s.

Explanation:

Given that

m= 1500 kg

r=60 m

θ = 18°

μ = 1

We know that maximum allowing speed for without slipping given as

$V=\sqrt{\dfrac{rg(tan\theta +\mu)}{1-\mu tan\theta}}$

Now by putting the values

$V=\sqrt{\dfrac{60\times 9.81(tan18^{\circ} +1)}{1- tan18^{\circ}}}$

V=33.98 m/s

The speed is 33.98 m/s.