Structures and Materials Question

[Dominatorrr]

A 16cm long animal tendon is found to extend 3.1mm by application of a force of 12.5N. Assume that the tendon is round with an average diameter of 8mm.

c) How much energy is stored in the tendon? Give your answer in J.

Wouldn't the answer just be Work OR Energy = Force * Change in length
Thus, W=12.5*0.0031= 0.03875 Joules

The answer in the back of my textbook says 0.019 J, exactly half of what I got. Why so?

[DisaFear]

Work is the amount of energy transferred into or out of the system, not exactly what you said.

Using the elastic potential formula U = 0.5kx^2 works in this case

Find the spring constant with Hooke's Law, using the values of force given and the extension, comes to around 4032 N/m
Then, using the elastic potential formula, you should get ~0.018N

Still confused as to what part the diameter plays in this. Where's this question from?

OH didn't notice this was a struc/mat question!!

[Dominatorrr]

Thanks a lot, you're totally right. Don't worry about the diameter, this is part c of the question, it has already been used in previous parts of the question. Oh and the question is from the Nelson 3/4 textbook, Exercise 6B, Question 3c.

OH didn't notice this was a struc/mat question!!

Look at the heading

[Dominatorrr]

Can someone show me how to do PART A of the question attached.

[Phy124]

Can someone show me how to do PART A of the question attached.

*Note M stands for moment (this is the same as torque, which I believe is used in year 12)

Taking g = 10 ms^-2



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[Dominatorrr]

Your working out seems legit, however the answer at the back of the Nelson Physics book says 1300N...hmm

[Aurelian]

Your working out seems legit, however the answer at the back of the Nelson Physics book says 1300N...hmm

So we have a cantilever of mass 6.0kg and length 3.0m. Assuming the cantilever is of uniform density, we can take the centre of mass to be at 1.5m. Thus, the torque applied by the cantilever itself (due to gravity) about the point where it meets the wall is given by t = F x d = 60N * 1.5m = 90Nm.

We also have a mass of weight 500N at the end of the cantilever. Thus, the torque due to this mass is given by t = F x d = 500N * 3.0m = 1500Nm.

Thus, the net torque in the clockwise direction is 1500Nm + 90Nm = 1590Nm. Therefore, the wire must apply an anticlockwise torque of equal magnitude in order to cancel out the clockwise torque.

Let us consider the vertical component of force applied by the wire on the cantilever. The wire joins the cantilever at 2.5m. That is, t = 1590Nm = Fvertical * 2.5m.

Rearranging gives Fvertical = 1590Nm/2.5m = 636N.

We can now use simple trigonometry (SOHCAHTOA) to find the tension in the wire; sin(30) = 636N/T  => T = 636N/sin(30) = 1272N = 1300N (to two significant figures).

Hope that helps!

[Phy124]

Oops, didn't read the question properly...

You need to include the weight of the cantilever.

This will act at the half way point of the beam (1.5m)





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edit: Ah, shit beaten by Aurelian, this is awkward 

[Dominatorrr]

Your working out seems legit, however the answer at the back of the Nelson Physics book says 1300N...hmm

So we have a cantilever of mass 6.0kg and length 3.0m. Assuming the cantilever is of uniform density, we can take the centre of mass to be at 1.5m. Thus, the torque applied by the cantilever itself (due to gravity) about the point where it meets the wall is given by t = F x d = 60N * 1.5m = 90Nm.

We also have a mass of weight 500N at the end of the cantilever. Thus, the torque due to this mass is given by t = F x d = 500N * 3.0m = 1500Nm.

Thus, the net torque in the clockwise direction is 1500Nm + 90Nm = 1590Nm. Therefore, the wire must apply an anticlockwise torque of equal magnitude in order to cancel out the clockwise torque.

Let us consider the vertical component of force applied by the wire on the cantilever. The wire joins the cantilever at 2.5m. That is, t = 1590Nm = Fvertical * 2.5m.

Rearranging gives Fvertical = 1590Nm/2.5m = 636N.

We can now use simple trigonometry (SOHCAHTOA) to find the tension in the wire; sin(30) = 636N/T  => T = 636N/sin(30) = 1272N = 1300N (to two significant figures).

Hope that helps!

Thanks! Really nice explanation.