Express modulus of elasticity in terms of 'mg'

Good day,

I think I've solved the question below, but I would like to double-check that my answer is correct.

I start with the following assumptions:

Let's imagine there's a vertical line splitting the triangle down the middle, into two right-angled triangles. We'll call it 'h'.

The extension of each string will be called 'x'.

We're told that tan(alpha) = 4 / 3.
=> h / a = 4 / 3
=> h = 4a / 3

The hypotenuse of this right triangle is a + x.
=> a + x = sqrt(a² + (4a / 3)²)
a + x = sqrt(a² + 16a²/ 9)
a + x = sqrt(25a²/ 9)
a + x = 5a / 3
x = (5a / 3) - a
x = 2a / 3

Now I'm calculating the tension in the strings using Hooke's Law. Since they're identical:

T = [lambda(2a/3)] / a
T = 2(lambda)/3

Resolving the forces vertically:

2Tsin(alpha) = mg
2[2(lambda)/3][h / (a+x)] = mg
[4(lambda)/3][(4a/3) / (5a/3)] = mg
[4(lambda)/3](0.8) = mg
16(lambda) / 15 = mg
lambda = 15mg / 16
lambda = 0.9375 mg

Would this be correct?

Thank you as always for your time and patience.

(edited 3 years ago)

Looks good.

Original post by mqb2766

Looks good.

Thank you once again. Mechanics is a particularly confusing topic for me for some reason.

Reply 3

mqb2766

Original post by DeadManProp

Thank you once again. Mechanics is a particularly confusing topic for me for some reason.

The only thing you could have done was to state the triangle was 3 : 4 : 5 so immediately the lengths are
a, 4a/3, 5a/3
Your basic pythagorean triplets often come up.

Reply 4

DeadManProp

Original post by mqb2766

The only thing you could have done was to state the triangle was 3 : 4 : 5 so immediately the lengths are
a, 4a/3, 5a/3
Your basic pythagorean triplets often come up.

Hmm, good point. I'm not sure how to determine it's a 3:4:5 triangle without checking the lengths first though.

(edited 3 years ago)

Reply 5

mqb2766

Original post by DeadManProp

Hmm, good point. I'm not sure how to determine it's a 3:4:5 triangle without checking the lengths first though.

No, but scalings of 1.5,2,2.5 or 6,8,10 often occur, or multiples of
5 : 12 : 13
Or
7 : 24 : 25

Your triangle is rightangled and the opp and adj sides are 3 & 4 (tan). Not hard.

(edited 3 years ago)

Reply 6

DeadManProp

Original post by mqb2766

No, but scalings of 1.5,2,2.5 or 6,8,10 often occur, or multiples of
5 : 12 : 13
Or
7 : 24 : 25

Your triangle is rightangled and the opp and adj sides are 3 & 4 (tan). Not hard.

Ahhh I completely forgot that the tangent is given to me right away. I got confused, as there was another question I saw where it wasn't given and yet I was meant to assume it's 3:4:5.

Reply 7

mqb2766

Original post by DeadManProp

Ahhh I completely forgot that the tangent is given to me right away. I got confused, as there was another question I saw where it wasn't given and yet I was meant to assume it's 3:4:5.