Irregular beam physics homework

[MrShimizu]

Homework Statement

A 1750-N irregular beam is haning horizontally by its ends from the ceiling by two vertical wires (A and B), each 1.25m long and weighing 2.50N. The center of gravity of this beam is one third of the way along the beam from the end where wire A is attached. A) If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling? B) Which pulse arrives first?

Homework Equations

T_a + T_b = 1750
Sum of all forces = 0.
wave speed = sqrt(force/linear mass density)
C_g is L/3 from A and 2L/3 from B

The Attempt at a Solution

I really just have hit a logical block and have no idea what to do.
------
L = 1.25
m = .255
mu = mass/length = .204
To find the force I tried: T = F * L/3
with F = 1750, a guess, really
1.25*1750/3 = 729.1666

Now I can't decide what to do with that number or even if it is a correct answer.

 

[Doc Al]

L = 1.25
m = .255
mu = mass/length = .204

OK.

To find the force I tried: T = F * L/3
with F = 1750, a guess, really

Don't guess. Set the net torque on the beam to zero and solve for the two tensions.

1.25*1750/3 = 729.1666

Not sure what you're doing here. Why did you multiply by 1.25?

Once you find the force, use it and mu to find the speed of the pulse. How long does each pulse take to travel to the ceiling?

 

[MrShimizu]

Don't guess. Set the net torque on the beam to zero and solve for the two tensions.

Such that:
F₁L₁-F₂L₂=0

L1 = 1.25/3 = .417
L2 = 2*1.25/3 = .833

?

 

[Doc Al]

Such that:
F₁L₁-F₂L₂=0

Yes.

L1 = 1.25/3 = .417
L2 = 2*1.25/3 = .833

No, the 1.25 is the length of the strings, not the length of the beam. Call the length of the beam "L". Then what are L₁ and L₂ in terms of L? You don't need the actual length of the beam--which isn't given--to solve for the forces.

 

[MrShimizu]

No, the 1.25 is the length of the strings, not the length of the beam.

This made me feel particularly... obtuse.

L₁ = L/3
L₂ = 2L/3

Then:
F₁L₁ - F₂L₂ = 0
(F₁*L)/3 - (F₂*2L)/3 = 0

Should F₁+F₂ = -1750?

 

[Doc Al]

L₁ = L/3
L₂ = 2L/3

Then:
F₁L₁ - F₂L₂ = 0
(F₁*L)/3 - (F₂*2L)/3 = 0

Good. Keep simplifying this. How does F₁ relate to F₂?

Should F₁+F₂ = -1750?

Yes, but get rid of the minus sign. Then you can solve for the forces by combining the two equations.

 

[MrShimizu]

F₁ = 2F₂
I got that from isolating F₁ in the earlier equation.

By plugging that into the lower equation

F₁=1166.67
F₂=583.33

v = sqrt(f/mu)

v₁=75.62m/s
v₂=53.46m/s

t=d/v;d=1.25m

t₁=.017s
t₂=.023s

Delta_t = .006s
The wave on String A arrives first.

 

[Doc Al]

F₁ = 2F₂
I got that from isolating F₁ in the earlier equation.

By plugging that into the lower equation

F₁=1166.67
F₂=583.33

v = sqrt(f/mu)

v₁=75.62m/s
v₂=53.46m/s

t=d/v;d=1.25m

t₁=.017s
t₂=.023s

Delta_t = .006s
The wave on String A arrives first.

Excellent!

My only suggestion would be to not round off to two significant figures until the end. (Use 3 sig figs when you calculate the time--that will give you a slightly different Delta t.)

 

[MrShimizu]

Thank you very much for your help, good Sir.

I'll also modify my answer as you advised.

Thanks again!