What is the tension in the string?

[risepj]

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_1_x+T_2_x+F_g_x=ma_x

T_1_y+T_2_y+F_g_y = T_1_y+T_2_y- F_g = ma_x

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_1 and T_2 as follows:

T_1_x+T_2_x+F_g_x=ma_x.

⇔ T_1cos(36.87°) - T_2cos(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_2 = T_1cos(36.87°)/cos(36.87°) = T_1.

T_1_y+T_2_y+F_g_y=0.

⇔ T_1sin(36.87°) +T_2sin(36.87°)=F_g

⇔ 2T_1sin(36.87°)=F_g

Therefore,

⇔ T_1=.5kg(9.8m/(s²))/(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_1+T_2 = 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!

 

[ehild]

⇔ T_1=.5kg(9.8m/(s²))/(2*sin(36.87°)) = 4.83N.

I think you just miss a zero. It is 4.083 N

ehild

 

[risepj]

Whoops, I think I actually did use 4.083 when I submitted my answer. Later in my post you see that I said that T_1+T_2 equals 8.16/8.17, which is double 4.08.

Editing first post to reflect that.