Question

a 0.16 kg meter - stick is held perpendicular to a vertical wall by a 2.5 m string going from the wall to the far end of the stick. find the tension in the string.

a. 0.86 n
b. 0.62 n
c. 0.92 n
d. 0.44 n
e. 0.33 n

Explanation:

Step1: Set up torque - equilibrium equation

Let the length of the meter - stick be $L = 1m$, mass $m=0.16kg$, and the length of the string be $r = 2.5m$. Take the point of contact of the meter - stick with the wall as the pivot. The weight of the meter - stick acts at its center of mass, which is at $L/2$ from the pivot. The weight of the meter - stick is $W=mg$, where $g = 9.8m/s^{2}$. The perpendicular distance from the line of action of the weight to the pivot is $L/2$, and the perpendicular distance from the line of action of the tension force $T$ to the pivot is $d=\sqrt{r^{2}-L^{2}}$ (using the Pythagorean theorem). In the equilibrium condition, the sum of torques about the pivot point is zero, $\sum\tau=0$. The torque due to the weight is $\tau_{W}=mg\times\frac{L}{2}$ (clock - wise), and the torque due to the tension is $\tau_{T}=T\times\sqrt{r^{2}-L^{2}}$ (counter - clockwise). So, $mg\times\frac{L}{2}=T\times\sqrt{r^{2}-L^{2}}$.

Step2: Solve for tension $T$

We know that $m = 0.16kg$, $g=9.8m/s^{2}$, $L = 1m$, and $r = 2.5m$. First, calculate $\sqrt{r^{2}-L^{2}}=\sqrt{(2.5)^{2}-1^{2}}=\sqrt{6.25 - 1}=\sqrt{5.25}\approx2.29m$. Then, from $mg\times\frac{L}{2}=T\times\sqrt{r^{2}-L^{2}}$, we can express $T$ as $T=\frac{mgL}{2\sqrt{r^{2}-L^{2}}}$. Substitute the values: $T=\frac{0.16\times9.8\times1}{2\times2.29}=\frac{1.568}{4.58}\approx0.33N$.

Answer:

e. 0.33 N