# Tension in a string

#### snormanlol

Homework Statement
A mass M (15kg) is hanging on a string which is forming a closed loop (total length of the loop = 5m, with mass per unit length=2*10^(-3) kg/m). The string runs trough 2 mass and frictionless wheels, with a distance of 2 meters between them. Determine the tension T in the string.
Homework Equations
F = m*a;
τ=I*α
I know that the answer has to be 98.6N. So I know that Fy=0 so that 2*T*sin(theta) = 147.25. Then I was think to take the torque of the left wheel but I cant find the lever arm of the tension force. I also know that u can solve the question by saying that the 2 sides of unknown length are 1.5 m but I'm not sure why u can do that.
Here is a picture of the problem:

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#### Delta2

Homework Helper
Gold Member
well I think this equation $2T\sin\theta=147.25$ is correct. All you need is a geometric argument to determine $\theta$. From the figure it seems that we can take the triangle to be isosceles. So how can you find $\theta$ if you know that the triangle is isosceles, has perimeter 5m and base 2m?

#### snormanlol

Well we can take Arctan(1/1.5) but I don't see why we can say that the perimeter is 5. Isn't the loop 5 m? And the length of the sides are unknown? That's why I tried using torque but I got the equation 2*T*sin(theta)=147.25 back.

#### Delta2

Homework Helper
Gold Member
What do you mean , since the total loop length is 5m, cant we say that the perimeter of the triangle is 5m?
Btw that should be $\theta=\arccos(1/1.5)$..

#### haruspex

Homework Helper
Gold Member
2018 Award
Isn't the loop 5 m?
Maybe the diagram, with its waveform in the horizontal section, is misleading you. Ignore the waveform at this stage (that is presumably for a later part of the question) and treat the horizontal part as straight. So the loop of string forms a triangle, and the length of the loop is the perimeter.

#### snormanlol

Ah yes it misleaded me thank you guys and yeah I meant arccos but I made a mistake. But thank you for the help guys.

"Tension in a string"

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