Vehicle shock absorbers damp out oscillations caused by road roughness. Describe how a temperature change may affect the operation of a shock absorber.

Short Answer

Expert verified
A temperature change can affect the operation of a shock absorber by changing the viscosity of the fluid within it. Warmer temperatures decrease viscosity, potentially making the shock absorber less effective. Conversely, colder temperatures increase viscosity, potentially making the shock absorber stiffer and less responsive.

Step by step solution

01

Operation of Shock Absorbers

Shock absorbers are mechanical or hydraulic devices designed to absorb and damp shock impulses in vehicles. They do this by converting the kinetic energy of the shock into another form of energy – typically heat – which is then dissipated.
02

Effect of Temperature on Viscosity

The basic working principle of a shock absorber involves a piston displacing fluid in a cylinder. This fluid has a property known as viscosity, a measure of a fluid's resistance to flow or deformation. The viscosity of fluids decreases as temperature increases. Thus, in normal conditions, the shock absorber operates at an optimal viscosity.
03

Resulting Effects on Shock Absorbers

When temperature increases, the lowered viscosity means that the fluid flows more freely. This in turn could cause the shock absorber to be less effective as it can not dissipate the same amount of energy as before. Conversely, when the temperature drops, the viscosity increases, meaning the fluid doesn't flow as easily. This could make the shock absorber stiffer and potentially less responsive. Therefore, temperature variations can significantly affect the operation of a shock absorber.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The information on a can of pop indicates that the can contains \(355 \mathrm{mL}\). The mass of a full can of pop is \(0.369 \mathrm{kg}\), while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at \(20^{\circ} \mathrm{C}\). Express your results in SI units.

The volume rate of flow, \(Q,\) through a pipe containing a slowly moving liquid is given by the equation \\[ Q=\frac{\pi R^{4} \Delta p}{8 \mu \ell} \\] where \(R\) is the pipe radius, \(\Delta p\) the pressure drop along the pipe, \(\mu\) a fluid property called viscosity \(\left(F L^{-2} T\right),\) and \(\ell\) the length of pipe. What are the dimensions of the constant \(\pi / 8 ?\) Would you classify this equation as a general homogeneous equation? Explain.

1.109 Air enters the converging nozzle shown in Fig. P1.72 at \(T_{1}=70^{\circ} \mathrm{F}\) ard \(V_{1}=50 \mathrm{ft} / \mathrm{s} .\) At the exit of the nozzle, \(V_{2}\) is given by \\[ V_{2}=\sqrt{V_{1}^{2}+2 c_{p}\left(T_{1}-T_{2}\right)} \\] where \(c_{p}=187 \mathrm{ft} \cdot \mathrm{lb} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\) and \(T_{2}\) is the air temperature at the exit of the nozzle, Find the temperature \(T_{2}\) for which \(V_{2}=\) \\[ 1000 \mathrm{ftfs} \\]

The density of a certain type of jet fuel is $775 \mathrm{kg} / \mathrm{m}^{3}$. Determine its specific gravity and specific weight.

A rigid-walled cubical container is completely filled with water at \(40^{\circ} \mathrm{F}\) and sealed. The water is then heated to \(100^{\circ} \mathrm{F}\) Determine the pressure that develops in the container when the water reaches this higher temperature. Assume that the volume of the container remains constant and the value of the bulk modulus of the water remains constant and equal to 300,000 psi.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free