# A curve with a radius of 125 m is properly banked for a car traveling 40 m/s. What must be the coefficient of static friction (µs) for a car not to skid on the same curve when traveling at 53 m/s?

\frac{1209}{1960} \approx .61
0

For the banked curve, the centripetal force necessary to make the turn is provided by the horizontal component of the normal force. When a car travels faster than the design speed, additional centripetal force must be provided by friction.

The centripetal force required for a car traveling at speed v on a curve of radius R is F_c = \frac{mv^2}{R}. The component of the normal force providing this centripetal force on a banked curve without friction is F_{c0} = \frac{mv_0^2}{R}.

The additional centripetal force required due to the increased speed must be provided by static friction f_s, which is \mu_s N, where N is the normal force and \mu_s is the coefficient of static friction.

Step Formula / Derivation Reasoning
1 F_{c0} = \frac{mv_0^2}{R} Centripetal force for design speed.
2 F_{c} = \frac{mv^2}{R} Centripetal force for actual speed.
3 f_s = F_{c} – F_{c0} Additional force required from static friction.
4 f_s = \mu_s N Static friction force.
5 N = mg \cos(\theta) Normal force on banked curve, where \theta is the banking angle.
6 \tan(\theta) = \frac{v_0^2}{Rg} Relationship for a banked curve without friction.
7 \mu_s = \frac{f_s}{N} Coefficient of static friction.
8 \mu_s = \frac{F_{c} – F_{c0}}{mg \cos(\theta)} Combine steps 3, 4, and 5.
9 \mu_s = \frac{\frac{mv^2}{R} – \frac{mv_0^2}{R}}{mg \cos(\theta)} Substitute F_{c} and F_{c0} from steps 1 and 2.
10 \mu_s = \frac{v^2 – v_0^2}{Rg \cos(\theta)} Mass m cancels out.
11 \mu_s = \frac{53^2 – 40^2}{125 \cdot 9.8 \cdot \cos(\theta)} Substitute values for v, v_0, and R.
12 \mu_s = \frac{2809 – 1600}{1225 \cdot 9.8 \cdot \cos(\theta)} Square the speeds.
13 \mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \cos(\theta)} Subtract the squares.
14 \mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \frac{40^2}{125 \cdot 9.8}} Substitute \cos(\theta) from step 6.
15 \mu_s = \frac{1209 \cdot 125}{1225 \cdot 40^2} Simplify.
16 \boxed{\mu_s = \frac{1209}{1960}} Final calculation for \mu_s.

The coefficient of static friction required is approximately \frac{1209}{1960}, which can be simplified further if needed.

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\frac{1209}{1960} \approx .61

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
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