The brake pads for a bicycle tire are made of

Oberlaudacu

Oberlaudacu

Answered question

2022-01-11

The brake pads for a bicycle tire are made of rubber.
If a frictional force of 50 N is applied to each side of the tires, determine the average shear strain in the rubber. Each pad has cross-sectional dimensions of 20 mm and 50 mm.
Gr=0.20MPa

Answer & Explanation

Jenny Sheppard

Jenny Sheppard

Beginner2022-01-12Added 35 answers

The base is 20 mm and 50 mm in size. The strain in the tire must be ascertained.Friction force
Ft=50N
Using the expressed shear stress, we will determine T.
Surface cross-sectional area:
A=5020
=1000mm2
τ=FtA
=501000
τ=0.05MPa
We can also calculate the shear stress in the following way.
τ=γG
Where:
γ- shear strain
τ- shear strain
γ=τG
=0.050.20
γ=0.25rad
The solution is:
γ=0.25rad

Shannon Hodgkinson

Shannon Hodgkinson

Beginner2022-01-13Added 34 answers

Given that Area of cross section (Acs)=50×20
Acs=1000mm2
Forse (F)=50N, G=02MPa
ShearstressPSK(τ)=FAcs=501000=0.05MPa
As we know Shearstra∈=ShearstressShearmodes
γ=τ4
γ=0.050.2=0.25rad
nick1337

nick1337

Expert2023-05-22Added 777 answers

Answer:
γ=0.00025
Explanation:
γ=ΔxL
where:
γ is the shear strain,
Δx is the change in length,
and L is the original length.
Given that the pads have cross-sectional dimensions of 20 mm and 50 mm, we can assume that the original length L is 20 mm. Since the pads are made of rubber, we can assume that the shear modulus of rubber (Gr) is 0.20 MPa.
The shear stress (τ) is given by:
τ=FA
where:
τ is the shear stress,
F is the applied force, and
A is the cross-sectional area.
In this case, the applied force (F) is 50 N, and the cross-sectional area (A) can be calculated as follows:
A={width}×{height}
Plugging in the values, we have:
A=20mm×50mm
Now, we can substitute the values into the formula for shear stress to find the shear strain:
τ=Gr·γ
Solving for γ, we get:
γ=τGr
Plugging in the values for τ and Gr, we have:
γ=50N20mm×50mm·1MPa0.20MPa
Now, we can simplify the equation:
γ=50N1000mm2·5
Converting the units to meters, we have:
γ=50N1000000mm2·5
Simplifying further:
γ=50×51000000
Thus, the average shear strain in the rubber is 0.00025.
Therefore, the average shear strain in the rubber is γ=0.00025.
Vasquez

Vasquez

Expert2023-05-22Added 669 answers

To solve the problem, we can use the equation for shear strain, which is given by:
γ=shear stressshear modulus
Given that the shear stress is equal to the frictional force applied on each side of the tire (50 N), and the shear modulus is given as G_r = 0.20 MPa, we can substitute these values into the equation to find the shear strain.
First, let's convert the shear modulus from MPa to N/m²:
Gr=0.20×106Pa
The cross-sectional area of each pad can be calculated as follows:
A={width}×{height}=(20mm)×(50mm)
Next, we can calculate the shear strain using the formula mentioned earlier:
γ=shear stressshear modulus=50N0.20×106Pa=500.20×106m/m
Finally, we can simplify the units by noting that 1 mm = 0.001 m:
γ=500.20×106×10.001=500.20×106×0.001m/m
Therefore, the average shear strain in the rubber is 500.20×106×0.001m/m.
RizerMix

RizerMix

Expert2023-05-22Added 656 answers

To determine the average shear strain in the rubber brake pads of a bicycle tire, we can use the formula:
{Shearstrain}=Shear stressShear modulus
Given:
- Shear stress, τ=50N
- Cross-sectional dimensions: d1=20mm (width) and d2=50mm (height)
- Shear modulus, Gr=0.20MPa=0.20×106N/m2
First, let's convert the dimensions to meters:
d1=20mm=20×103m
d2=50mm=50×103m
Now, we can calculate the average shear strain:
{Shearstrain}=Shear stressShear modulus
Substituting the given values:
{Shearstrain}=50N0.20×106N/m2
Simplifying the expression:
{Shearstrain}=500.20×106=0.25×103
Therefore, the average shear strain in the rubber brake pads is 0.25×103 rad.

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