the figure below shows a horizontal rod, of length 6.2 m, with forces acting on it. a 30 n force acts at its left end, point o, in a direction down and to the left, 45° below horizontal. a 25 n force acts at its center, point c, in a direction up and to the right, 30° to the right of vertical. a 10 n force acts at the right end, in a direction down and to the right, 20° below horizontal.
(a) what is the net torque (in n - m) on the rod about an axis through o perpendicular to the page?
magnitude n - m
direction
(b) what is the net torque (in n - m) on the rod about an axis through c perpendicular to the page?
magnitude n - m
direction

Question

Accepted Answer

# Explanation:
## Step1: Recall the torque formula
Torque is given by $\tau = rF\sin\theta$, where $r$ is the distance from the axis of rotation, $F$ is the force, and $\theta$ is the angle between the position vector and the force vector.

## Step2: Calculate torque for each force about axis through $O$
- For the $30 - N$ force: The line of action of the $30 - N$ force passes through $O$, so $r = 0$ and $\tau_{30}=0$
- For the $25 - N$ force: $r = 3.1\space m$, $F = 25\space N$, $\theta=60^{\circ}$ (since the angle with the vertical is $30^{\circ}$, the angle with the horizontal - perpendicular to the rod - is $60^{\circ}$). Then $\tau_{25}=3.1\times25\times\sin60^{\circ}$
$$
\begin{align*}
\tau_{25}&=3.1\times25\times\frac{\sqrt{3}}{2}\
&=3.1\times25\times0.866\
& = 67.2\space N\cdot m
\end{align*}
$$
- For the $10 - N$ force: $r = 6.2\space m$, $F = 10\space N$, $\theta = 20^{\circ}$. Then $\tau_{10}=6.2\times10\times\sin20^{\circ}$
$$
\begin{align*}
\tau_{10}&=6.2\times10\times0.342\
&=21.2\space N\cdot m
\end{align*}
$$
- Net torque about $O$: $\tau_{net,O}=\tau_{25}-\tau_{10}$ (assuming $\tau_{25}$ is counter - clockwise and $\tau_{10}$ is clockwise)
$$
\begin{align*}
\tau_{net,O}&=67.2 - 21.2\
&=46\space N\cdot m
\end{align*}
$$

## Step3: Calculate torque for each force about axis through $C$
- For the $30 - N$ force: $r = 3.1\space m$, $F = 30\space N$, $\theta = 45^{\circ}$. Then $\tau_{30}=3.1\times30\times\sin45^{\circ}$
$$
\begin{align*}
\tau_{30}&=3.1\times30\times\frac{\sqrt{2}}{2}\
&=3.1\times30\times0.707\
&=66.7\space N\cdot m
\end{align*}
$$
- For the $25 - N$ force: The line of action of the $25 - N$ force passes through $C$, so $r = 0$ and $\tau_{25}=0$
- For the $10 - N$ force: $r = 3.1\space m$, $F = 10\space N$, $\theta = 20^{\circ}$. Then $\tau_{10}=3.1\times10\times\sin20^{\circ}$
$$
\begin{align*}
\tau_{10}&=3.1\times10\times0.342\
&=10.6\space N\cdot m
\end{align*}
$$
- Net torque about $C$: $\tau_{net,C}=\tau_{10}-\tau_{30}$ (assuming $\tau_{30}$ is clockwise and $\tau_{10}$ is counter - clockwise)
$$
\begin{align*}
\tau_{net,C}&=10.6-66.7\
&=- 56.1\space N\cdot m\
|\tau_{net,C}|&=56.1\space N\cdot m
\end{align*}
$$

# Answer:
(a) magnitude: $46\space N\cdot m$, direction: counter - clockwise
(b) magnitude: $56.1\space N\cdot m$, direction: clockwise

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the torque formula

Step2: Calculate torque for each force about axis through \(O\)

Step3: Calculate torque for each force about axis through \(C\)

Answer: