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3 years ago

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The figure shows the arrangement of master cylinder and slave cylinder of a part of braking system. The master cylinder piston,

of surface area 8 cm2 exerts a force of 28 N on the brake fluid when the driver presses the brake pedal Calculate the pressure on the brake fluid produced by the master piston This pressure is transmitted to the slave piston that has surface area of 40 cm2. Calculate the force exerted on this piston State two methods of increasing the force produced by the slave cylinder piston.

1 answer:

Neko [114]3 years ago

4 0

Answer:

a) 35 kPa

d) 140 N

c) 1) Increasing the brake fluid pressure

2) Increasing the slave piston surface area.

Explanation:

The parameters given are;

a) Force applied to the master cylinder piston = 28 N

Cross sectional area of the master cylinder piston = 8 cm² = 8 × 10⁻⁴ m²

The pressure P on the brake fluid is given by the formula for pressure as follows;

$P= \dfrac{Applied \ force}{Area \over \ which \ force \ is \ applied} = \dfrac{28 \, N}{8 \times 10^{-4} \, m^2} = 35,000 \, N/m^2 = 35,000 \, Pa$

The pressure on the brake fluid, P, produced by the master cylinder piston = 35,000 Pa = 35 kPa

b) Given that the area of the slave piston = 40 cm² = 0.004 m², we have from the formula for pressure, P;

$P= \dfrac{Applied \ force}{Area \over \ which \ force \ is \ applied} = \dfrac{Applied \ force}{4 \times 10^{-3} \, m^2} = 35,000 \, N/m^2$

Therefore;

Applied force on the slave piston = 4 × 10⁻³ m² × 35,000 N/m² = 140 N

c) The force, F produced by the slave cylinder piston is given by the relation;

F = Pressure × Area

Therefore, the two ways of increasing the force produced by the slave cylinder piston is as follows;

1) Increasing the pressure in the brake fluid by increasing the force exerted by the master cylinder piston

2) Increasing the surface area of the slave piston.

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A) The velocity of sphere A after the collision is 1.00 m/s to the right

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C) The velocity of sphere C is 2.68 m/s at a direction of $-5.2^{\circ}$

D) The impulse exerted on C is 4.29 kg m/s at a direction of $-5.2^{\circ}$

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

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$v_A$ is the final velocity of sphere A

$m_B = 1.80 kg$ is the mass of sphere B

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B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

$K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J$

After the collision:

$K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J$

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

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While the total momentum along the y-axis is zero:

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D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

$p_C = 0$

While the final momentum is:

$p'_C = 4.29 kg m/s$

So the magnitude of the impulse exerted on C is

$I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s$

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore $-5.2^{\circ}$ .

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

$K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J$

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

$K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J$

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

$F=Ma$

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

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