# Chapter 9 - Rotational Dynamics - Problems - Page 244: 9

$\theta=43.7^o$

#### Work Step by Step

Let's call the distance from the axis of rotation to the point where 2 forces are applied $l$. - The first force $F_1=38N$. Since it is applied perpendicular to the rod, the lever arm is directly $l$. Therefore, $$\tau_1=38l$$ - The second force $F_2=55N$. Since it is directed at angle $\theta$, the component which is perpendicular with the rod, it has $l$ as the lever arm, and the force is $F_2\sin\theta=55\sin\theta$. $$\tau_2=55l\sin\theta$$ Since the net torque is zero, we take $\tau_1$ to be negative, so $$\tau_2-\tau_1=0$$ $$55l\sin\theta=38l$$ $$\sin\theta=\frac{38}{55}$$ $$\theta=43.7^o$$

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