Question

A 1500kg car is placed on the large piston with a radius of 4 m i. what is the maximum upward force that the large piston must exert to lift the car? ii. If the mechanical advantage of the hydraulic system is 20, what is the minimum downward force that should be applied to lift the car? iii. What is the radius of the small piston? iv. How far should the small piston descend in order to lift the car by 2 m?

Answer 1

Answer:

In a hydraulic lift, used at a service station the radius of the large and small piston are in the ratio of 20:1. What weight placed on the small piston will be sufficient to lift a car of mass 1500kg ?

A

3.75kg

B

37.5kg

C

7.5kg

D

75kg

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ANSWER

Answer is A.

Pressure is the amount of force acting per unit area. That is, P=F/A.

where:

p is the pressure,

F is the normal force,

A is the area of the surface on contact. Let us consider A = πr2.

Therefore, πr12F1=πr22F2.

In this case, 2021500=12W,W=3.75kg.

Hence, weight to be placed on the small piston sufficient to lift a car of mass 1500 kg is 3.75 kg.

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Answer 2

Answer:

For the requested velocity, a traction torque of $330 \mathrm{Nm}$ a rotational speed  $1004 \mathrm{rpm}$ is required. This is equivalent to the acceleration caused by a slope of$2.7 \% .$

Explanation:

Given: A 1500kg car is placed on a large piston with a radius of 4 m

To find: Torque, Speed, Acceleration

Solution:

Assume $\rho_{a}=1.20 \mathrm{~kg} / \mathrm{m}^{3}, g=9.81 \mathrm{~m} / \mathrm{s}^{2} .$

a) Traction torque required at wheels:

$\begin{aligned}F_{t} &=m_{v} \cdot c_{r} \cdot g+1 / 2 \cdot \rho_{a} \cdot A_{f} \cdot c_{d} \cdot v^{2}+m_{v} \cdot a=\\&=1500 \cdot 0.012 \cdot 9.81+\frac{1}{2} \cdot 1.2 \cdot 0.7 \cdot\left(\frac{120}{3.6}\right)^{2}+1500 \cdot 0.027 \cdot 9.81=1041 \mathrm{~N}\end{aligned}$

The information about the tires is explained by a newline

$\underbrace{195}_{\text {width of the }} / \underbrace{65}_{\text {ratio of sidewall }} \quad / \underbrace{15}_{\text {height to tire }} \quad \underbrace{\mathrm{T}}_{\text {wheel diameter }} \quad max. 190[\mathrm{~km} / \mathrm{h}]$

Thus

$\begin{aligned}r_{w} &=\frac{d_{w}}{2}+h_{\mathrm{sw}}=\frac{15 "}{2}+0.65 \cdot 0.195=15 \cdot \frac{0.0254}{2}+0.65 \cdot 0.195=0.317 \mathrm{~m} \\T_{t} &=r_{w} \cdot F_{t}=0.317 \cdot 1041=330 \mathrm{Nm}\end{aligned}$

b) Rotational speed level:

$\omega_{w}=\frac{v}{r_{w}}=\frac{120 / 3.6}{0.317} \cdot 0.317=105.2 \mathrm{rad} / \mathrm{s}=1004 \mathrm{rpm}$

c) Acceleration-equivalent road slope:

$\begin{aligned}\alpha &=\arcsin \left(\left(\frac{a}{g}=0.027\right) \mathrm{rad}\right.\\\alpha_{\%} &=100 \cdot \tan (0.027)=100 \cdot 0.027=2.7 \%\end{aligned}$

For the requested velocity, a traction torque of $330 \mathrm{Nm}$ at a rotational speed  $1004 \mathrm{rpm}$ is required. This is equivalent to the acceleration caused by a slope of$2.7 \% .$