- #1

risepj

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## Homework Statement

"What is the tension in the string?"

- .5kg block

- Suspended at the midpoint of a 1.25m-long string

- Ends of string attached to ceiling are 1m apart.

## Homework Equations

cosθ, sinθ, tanθ

T[itex]_1[/itex][itex]_x[/itex]+T[itex]_2[/itex][itex]_x[/itex]+F[itex]_g[/itex][itex]_x[/itex]=ma[itex]_x[/itex]

T[itex]_1[/itex][itex]_y[/itex]+T[itex]_2[/itex][itex]_y[/itex]+F[itex]_g[/itex][itex]_y[/itex] = T[itex]_1[/itex][itex]_y[/itex]+T[itex]_2[/itex][itex]_y[/itex]- F[itex]_g[/itex] = ma[itex]_x[/itex]

## The Attempt at a Solution

My homework is online and immediately verifies whether or not your answers are correct. I was able to solve that the angle the string makes with the ceiling is 36.87°.

From there, I attempted to solve for T[itex]_1[/itex] and T[itex]_2[/itex] as follows:

T[itex]_1[/itex][itex]_x[/itex]+T[itex]_2[/itex][itex]_x[/itex]+F[itex]_g[/itex][itex]_x[/itex]=ma[itex]_x[/itex].

⇔ T[itex]_1[/itex]cos(36.87°) - T[itex]_2[/itex]cos(36.87°) + 0 = 0 (since the force of gravity has no affect on the x-component of the tensions, and the object is not accelerating).

⇔ T[itex]_2[/itex] = T[itex]_1[/itex]cos(36.87°)/cos(36.87°) = T[itex]_1[/itex].

T[itex]_1[/itex][itex]_y[/itex]+T[itex]_2[/itex][itex]_y[/itex]+F[itex]_g[/itex][itex]_y[/itex]=0.

⇔ T[itex]_1[/itex]sin(36.87°) +T[itex]_2[/itex]sin(36.87°)=F[itex]_g[/itex]

⇔ 2T[itex]_1[/itex]sin(36.87°)=F[itex]_g[/itex]

Therefore,

⇔ T[itex]_1[/itex]=.5kg(9.8m/(s

^{2}))/(2*sin(36.87°)) = 4.083N.

I've used up all but one of my attempts. Maybe it's because I'm not understanding the wording of the problem. I used up two attempts by guessing T[itex]_1[/itex]+T[itex]_2[/itex] = 8.16/8.17 (I thought that it may be an issue with rounding).

Anyway, if someone could provide some input as to what the problem is asking, I would be extremely appreciative! And it would also be nice to have someone verify that my math and setup of this problem is correct. Thank you so much!

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