Ask question

Ask question

All categories

3 years ago

15

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

You might be interested in

You move a 4 N object 10 meters. Find work

Trava [24]

Answer: 40 J

Explanation: Work is equal to the product of force and distance.

W = Fd

= 4N ( 10 m)

= 40 J

8 0

3 years ago

A stone is thrown from a cliff and lands in the sea. Air resistance is negligible. Which statement is correct whilst the stone i

erik [133]

Since the stone is travelling, the displacement cannot be constant. This leaves the last 2 choices as possibilities. Now consider the acceleration that can act on the stone. Acceleration due to gravity is the only one. That means the horizontal component of the stone has to be constant since there is no acceleration in the horizontal direction.

4 0

4 years ago

A cue ball has a mass of 0.5 kg. During a game of pool, the cue ball is struck and now has a velocity of 3 . When it strikes a s

photoshop1234 [79]

Answer: 3 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:

$p_i = p_f$

The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore

$p_i = m_c u_c = (0.5 kg)(3 m/s)=1.5 kg m/s$

So, the final total momentum will also be

$p_f = 1.5 kg m/s$

And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:

$p_f = m_s v_s$

from which we find the velocity of the solid ball

$v_s = \frac{p_f}{m_s}=\frac{1.5 kg m/s}{0.5 kg}=3 m/s$

8 0

3 years ago

Read 2 more answers

What roles do the musk ox play in the tundra ecosystem? (Select all that apply.)

docker41 [41]

Answer:

it's pray hoped this helped

3 0

3 years ago

Read 2 more answers

Select the correct answer

Nadya [2.5K]

Answer:

$f= 3.0 \times 10 {}^{8} \div 7.0 \times 10 {}^{7} \\ f = 4.28hz$

Given

7 0

3 years ago