A simple friction stopping device is formed by pressing 2 elastic blocks against a metal spinning rim. The contact areabetween each block and metal rim is 0.05mx0.02m and the blocks have a thickness of 0.01m each. If the blocks have a1.shear modulus of 600 kPa and the friction force applied to each side of the rim is 0.05 kN determinea) The shear strain in each block Ans: 0.0833 radb) The deflection of the pad across the 10 mm thickness50 mm10 mm10 mm

Given data: The contact area between each block and the metal rim is 0.05 m ×0.02 m The thickness…

Answered: A simple friction stopping device is…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Question

A simple friction stopping device is formed by pressing 2 elastic blocks against a metal spinning rim. The contact area
between each block and metal rim is 0.05mx0.02m and the blocks have a thickness of 0.01m each. If the blocks have a
1.
shear modulus of 600 kPa and the friction force applied to each side of the rim is 0.05 kN determine
a) The shear strain in each block Ans: 0.0833 rad
b) The deflection of the pad across the 10 mm thickness
50 mm
10 mm
10 mm

Expert Solution

Step 1

Given data:

The contact area between each block and the metal rim is $0.05 m \times 0.02 m$
The thickness of the block is $0.01 m$
The shear modulus for the block is $G = 600 kPa$
Fiction force due to each side of the rim is $0.05 kN$

(a) The shear stress $τ$ can be expressed as,

$\begin{array}{rcl} τ & = & G \cdot γ \\ γ & = & \frac{τ}{G} \\ = & \frac{F}{G \cdot A} \end{array}$

Here, $G$ is the shear modulus then for the shear strain $γ$ ,

$\begin{array}{rcl} γ & = & \frac{0.05 kN}{600 kPa \times (0.05 m \times 0.02 m)} \\ = & \frac{0.05 kN \times \frac{10^{3} N}{1 kN}}{600 kPa \times \frac{10^{3} Pa}{1 kPa} \times (0.05 m \times 0.02 m)} \\ = & \frac{0.05 \times 10^{3} N}{600 \times 10^{3} Pa \times (0.05 \times 0.02) m^{2}} \\ = & \frac{0.05 \times 10^{3} N}{(600 \times 10^{3} \times 0.05 \times 0.02) Pa \cdot m^{2} \times \frac{1 N}{1 Pa \cdot m^{2}}} \\ = & 0.0833 rad \end{array}$

Thus, the shear strain in each block is $0.0833 rad$ .

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions