Question

current attempt in progress
one end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. two forces, both parallel
to the tabletop, are applied to the stick in such a way that the net torque is zero. the first force has a magnitude of 2.00 n and is applied
perpendicular to the length of the stick at the free end. the second force has a magnitude of 6.00 n and acts at a 27.4° angle with
respect to the length of the stick. where along the stick is the 6.00 - n force applied? express this distance with respect to the end of
the stick that is pinned.

d =

Explanation:

Step1: Recall the torque formula

Torque $\tau = rF\sin\theta$. For the first force $F_1 = 2.00\ N$, $r_1=1\ m$ (since it's at the free - end of the meter - stick) and $\theta_1 = 90^{\circ}$ (perpendicular), so $\tau_1=r_1F_1\sin\theta_1=(1\ m)\times(2.00\ N)\times\sin90^{\circ}=2.00\ N\cdot m$.

Step2: Analyze the second - force torque

Let the distance of the second force from the pinned end be $d$. The second force $F_2 = 6.00\ N$ and $\theta_2 = 90^{\circ}-27.4^{\circ}=62.6^{\circ}$. The torque due to the second force is $\tau_2=dF_2\sin\theta_2$.

Step3: Use the net - torque condition

Since $\tau_{net}=\tau_1-\tau_2 = 0$ (the torques must be equal in magnitude and opposite in direction), we have $r_1F_1\sin\theta_1=dF_2\sin\theta_2$.
Substitute the known values: $2.00\ N\cdot m=d\times(6.00\ N)\times\sin62.6^{\circ}$.
We know that $\sin62.6^{\circ}\approx0.891$. Then $d=\frac{2.00\ N\cdot m}{6.00\ N\times0.891}$.

Answer:

$d=\frac{2}{6\times0.891}\approx0.376\ m$