# Thread: Another static questing finding tension

1. ## Another static questing finding tension

[COLOR=#2D3B45]The diagram shows a model rod OQ of 5.72 kg mass. The upper end Q of the rod is attached a model string of length 1.34 m, whose other end P is attached to a fixed point on a vertical wall. The end O of the rod is hinged to the vertical wall. The length of the string is equal to the length of the rod. The angle α is 0.83 radians.

Calculate the tension in the string PQ, to 3 decimal places. Assume g = 9.81 m/s2.
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I would like to know if im doing it correctly. 1st, i drew a line across the middle OP, connecting it to Q and named the point where Q meets OP C. W=5.72*9.81 and the perpendicular distance from the pivot point to the line of action of the weight is (1.34/2)sin0.83rad. Therefore, clockwise torque=5.72*9.81*(1.34/2)sin0.83rad. OP=2OC. Therefore, OP= 2(1.34)cos0.83rad= 2.68cos0.83rad. The perpendicular distance from the pivot point to the line of action of the tension is 2.68cos0.83rad*sin0.83rad. Therefore, ccw torque= T*2.68cos0.83rad*sin0.83rad. Ccw torque=clockwise torque. Therefore, T= (5.72*9.81*(1.34/2)sin0.83rad)/(2.68cos0.83rad*sin0.83rad)=20.786N. Also, kindly help me check if my final answer is correct .

2. ## Re: Another static questing finding tension

opps...i typed the title wrongly LOL.rip...

3. ## Re: Another static questing finding tension

Your diagram is not showing. My interpretation of your written description ...

$\dfrac{mgL}{2} \cdot \sin{\alpha} = T \cdot L \cdot \sin(\pi-2\alpha)$

$\dfrac{mg}{2} \cdot \dfrac{\sin{\alpha}}{\sin(\pi-2\alpha)} = T$

substituting in your given values for $m$, $g$, and $\alpha$ yields your solution fot $T$.

4. ## Re: Another static questing finding tension

here's the diagram