Calculating the tesion in a string holding a 1 meter stick

[saurabheights]

A meterstick (L = 1 m) has a mass of m = 0.13 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.



What is the tension in the left string right after the right string is cut?



Calculating the free body diagram where we put the acceleration of center of mass of the plank, mg and T.

 

[Doc Al]

Hint: Consider torques about the point where the left string attaches to the stick.

 

[saurabheights]

Yeah did that. I first calculated torque about the left string using integral calculas over a thin piece of stick at distance x. The total torque calculated was MGL/4. Then calculated Inertia. It came out to be 7ML^2/48. Then, I calculated angular acceleration as = 7g/12L.

Finally, I calculated tension using free body diagram → T-mg+ ∫(dm * α * x) = 0
→T-mg+∫(M/L*α*x.dx) = 0
where x went from -0.25 to +0.75.

T = mg-∫(ML*dm/L*x.dx)

Is this the correct solution?

 

[Doc Al]

Yeah did that. I first calculated torque about the left string using integral calculas over a thin piece of stick at distance x. The total torque calculated was MGL/4. Then calculated Inertia. It came out to be 7ML^2/48.

Good.

Then, I calculated angular acceleration as = 7g/12L.

Redo that calculation.

Finally, I calculated tension using free body diagram → T-mg+ ∫(dm * α * x) = 0

Not sure what you are doing here or what that last term represents.

Just use: ƩF = ma

 

[Nickie]

I would approach this question by first identifying the forces acting on the meterstick. These include its weight (mg), the tension in the left string (T), and the tension in the right string (T').

Next, I would draw a free body diagram to represent these forces and their respective directions. Since the meterstick is initially at rest, we can assume that the acceleration of its center of mass is zero. Therefore, the net force acting on the meterstick must also be zero according to Newton's Second Law (F=ma).

From the diagram, we can see that the weight of the meterstick is acting downwards and is balanced by the tension in the left and right strings. Since the right string is cut, the tension in that string becomes zero and the weight of the meterstick is now only balanced by the tension in the left string.

Using Newton's Second Law, we can set up the following equation: ΣF = 0 = T - mg. Solving for T, we get T = mg = (0.13 kg)(9.8 m/s^2) = 1.274 N.

Therefore, the tension in the left string right after the right string is cut is approximately 1.274 N. It is important to note that this is only an approximation as there may be other factors such as the mass distribution of the meterstick and the elasticity of the strings that could affect the actual tension. Further experimentation and calculations would be needed to determine a more accurate value.