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3 years ago

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Consider a wet banked roadway, where there is a coefficient of static friction of 0.30 and a coefficient of kinetic friction of

0.25 between the tires and the roadway. The radius of the curve is R=50m.
Part A

If the banking angle is β=25∘, what is the maximum speed the automobile can have before sliding up the banking?

Part B

What is the minimum speed the automobile can have before sliding down the banking?

1 answer:

gulaghasi [49]3 years ago

6 0

Answer:

a) v = 20.9 m/s

b) v = 8.46 m/s

Explanation:

Given:-

- The coefficient of static friction is us = 0.30

- The coefficient of static friction is uk= 0.25

- The radius of the curve R = 50m

- The bank Angle β = 25

Find:-

a) If the banking angle is β=25∘, what is the maximum speed the automobile can have before sliding up the banking?

b) What is the minimum speed the automobile can have before sliding down the banking?

Solution:-

- We will investigate the sliding-up case first. Develop a FBD as given in (attachment).

- Use Newton's second law of motion vertical to slope of bank where the car is in equilibrium:

Sum ( F_n ) = 0

N*cos(β) - m*g - Ff*sin(β) = 0

Where, Frictional Force Ff = us*N

N (cos(β) - us*sin(β)) = mg ... Eq 1

- Use Newton's second law of motion horizontal to slope of bank where the car is accelerating:

Sum ( F_h ) = m*a

Ff*cos(β) + Nsin(β) = m*v^2 / R

N (us*cos (β) + sin (β) ) = m*v^2 / R .... Eq 2

- Divide the two equations:

v^2 / gR = [ us*cos (β) + sin (β) ] / [ cos (β) - us*sin (β) ]

v^2 = [ 0.25*cos (25) + sin (25) ]*9.81*50 / [ cos (25) - 0.25*sin (25) ]

v = 20.9 m/s

- For the slip down case. We have, friction force Ff reversed hence us = -us. Then the v can be given as:

v^2 / gR = [ -us*cos (β) + sin (β) ] / [ cos (β) + us*sin (β) ]

v^2 = [ -0.25*cos (25) + sin (25) ]*9.81*50 / [ cos (25) + 0.25*sin (25) ]

v = 8.46 m/s

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

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