Question

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 53.7° 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.

Explanation:

Step1: Recall the torque formula

Torque $\tau = rF\sin\theta$. For the first force, $r_1 = 1\space m$ (length of the meter - stick), $F_1=2.00\space N$, $\theta_1 = 90^{\circ}$, so $\tau_1=r_1F_1\sin\theta_1=(1\space m)\times(2.00\space N)\times\sin90^{\circ}=2.00\space N\cdot m$.
For the second force, let the distance from the pinned - end be $d$, $F_2 = 6.00\space N$, $\theta_2=53.7^{\circ}$, so $\tau_2=dF_2\sin\theta_2$.

Step2: Use the condition of zero net torque

Since $\tau_{net}=\tau_1-\tau_2 = 0$ (the torques are in opposite directions to cancel each other), we have $\tau_1=\tau_2$.
Substitute the expressions for $\tau_1$ and $\tau_2$:

$$
LATEXBLOCK0
$$

$$ d=\frac{2.00\space N\cdot m}{6.00\space N\times0.806} $$

$$ d=\frac{2.00}{4.836}\space m\approx0.414\space m $$

Answer:

$d = 0.414\space m$