How Do You Calculate Tractive Force for Accelerating a Car?

[Tom.G]

I think it was Sir Isaac Newton that stated something like this:
"A body at rest tends to stay at rest."
"A body in motion tends to stay in motion."

Since the body you are talking about is a car that is trying to accelerate, its tendency (inertia) is tending to keep it at rest.

I know, we usually use the word "inertia" in the case of something that is moving tends to keep moving; but it works both ways!

So for a FBD, the arrow for Inertia points toward where the object (car) would be if no force were applied to it.

Hope this helps!

Cheers,
Tom

 

[Gixer1127]

I think it was Sir Isaac Newton that stated something like this:
"A body at rest tends to stay at rest."
"A body in motion tends to stay in motion."

Since the body you are talking about is a car that is trying to accelerate, its tendency (inertia) is tending to keep it at rest.

I know, we usually use the word "inertia" in the case of something that is moving tends to keep moving; but it works both ways!

So for a FBD, the arrow for Inertia points toward where the object (car) would be if no force were applied to it.

Hope this helps!

Cheers,
Tom

Thanks Tom. Yup, that's what I'm trying to get my head around. I'll get there in the end and have almost completed this particular assessment with the help of the folks on here. Very much obliged to all.

 

[Gixer1127]

Hey folks. I've just had this come back as wrong from my tutor. He's saying the decimal point in (a) is in the wrong place. This in turn leads onto an incorrect answer for (b) and (c) Any pointers on where I've gone wrong?

 

[Gixer1127]

View attachment 325369View attachment 325369
Hey folks. I've just had this come back as wrong from my tutor. He's saying the decimal point in (a) is in the wrong place. This in turn leads onto an incorrect answer for (b) and (c) Any pointers on where I've gone wrong?

It's OK, I've seen the error of my ways

 

[Lnewqban]

... Hey folks. I've just had this come back as wrong from my tutor. He's saying the decimal point in (a) is in the wrong place. This in turn leads onto an incorrect answer for (b) and (c) Any pointers on where I've gone wrong?

In your calculations, avoid using mm for length, area or volume.
Before you start plugging numbers into equations, transfer all given data to the International System of Units:

https://en.wikipedia.org/wiki/International_System_of_Units

In your case, make those 16 mm of diameter 16/1000 meters first.
One good clue about your error in a) answer was the result of a fraction of 1 mm² as the cross-section area of a circle of 16 mm diameter: impossible.

Another clue on b) and d) answers: Pascals are N/m² rather than N/mm².

 

[Gixer1127]

(b) Tensile strain (ɛ) = σ/E

Modulus of elasticity (E) of the bolt is 160GPa.

ɛ = 34468656.72Pa / 160x10⁹GPa = 0.0002154 (2.154x10¯⁴) (No units of measurement for strain)

(c) Extension = ɛ x original length

Extension = 0.0002154 x 20mm = 0.004308mm(4.308x10¯³)

I'm thinking the above in bold is not correct? Should I be converting the 20mm to 0.02m and working it that way? In which case the answer should be 4.308x10¯⁶ (0.000004308mm).

 

[Gixer1127]

(b) Tensile strain (ɛ) = σ/E

Modulus of elasticity (E) of the bolt is 160GPa.

ɛ = 34468656.72Pa / 160x10⁹GPa = 0.0002154 (2.154x10¯⁴) (No units of measurement for strain)

(c) Extension = ɛ x original length

Extension = 0.0002154 x 20mm = 0.004308mm(4.308x10¯³)

I'm thinking the above in bold is not correct? Should I be converting the 20mm to 0.02m and working it that way? In which case the answer should be 4.308x10¯⁶ (0.000004308mm).

Nope, going with 20mm, not 0.02m.

 

[Lnewqban]

Nope, going with 20mm, not 0.02m.

Where those 20 mm come from?
34468656.72Pa may be the wrong amount, based on the erroneously calculated cross-section area.
160x10⁹ GPa is not correct; it is either 160x10⁹ Pa or 160 GPa.

 

[Gixer1127]

Sorry, yes, I missed out in my original post where the 20mm comes into play.
The 16 mm diameter steel bolt with modulus of elasticity of 160 GPa and modulus of rigidity of 90 GPa shown in the diagram below holds two components together. The thickness of the bolted interface of the components is 20 mm.
And the question reads: (c) Calculate the change in length of the 20mm bolted interface of the bolt.
I'll change, in my calculation summary, to 160x10⁹ Pa, thank you.
I'm going to stick to my original answer, albeit with the correct decimal point placement. (34468656.72Pa) In my original calculation I had not input the data correctly in my calculator which gave me an incorrect decimal point placement.
So here's my final answers below with the 3 questions I got wrong due to my first answer being incorrect.

Calculate the direct stress in the bolt at the centre of the joined components. Calculate the tensile strain experienced by the bolt. Calculate the change in length of the 20mm bolted interface of the bolt.

Area of cross section of 16mm dia. bolt is.
A = πr² = π x 0.008² = 2.01x10¯⁴mm² (0.000201mm²)

(a)Direct stress (σ) = Force/Area

= 6928.2N / 2.01(10¯⁴)mm² = 34468656.72 Pa (0.3446865672x10⁸)

(b) Tensile strain (ɛ) = σ/E

Modulus of elasticity (E) of the bolt is 160GPa.

ɛ = 34468656.72Pa / 160x10⁹Pa = 0.0002153 (2.153x10¯⁴) (No units of measurement for strain)

(c) Extension = ɛ x original length

Extension = 0.0002153 x 20mm = 0.004306mm(4.306x10¯³mm)

 

[Tom.G]

Area of cross section of 16mm dia. bolt is.
A = πr² = π x 0.008² = 2.01x10¯⁴mm² (0.000201mm²)

HUH?? How can the area be less than the diameter?

 

[Gixer1127]

Hi Tom. I thought I'd posted a reply a couple of days ago but obviously didn't push the send button.
You are 100% correct, of course. I used the wrong unit of measurement so thank you for spotting this. I have changed this and submitted for marking.