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Expert-verified Found in: Page 382 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # A uniform plank of length 2.00 m and mass 30.0 kg is supported by three ropes as indicated by the blue vectors in Figure P12.25. Find the tension in each rope when a 700-N person is ${\mathbit{d}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{500}}{\mathbit{m}}$ from the left end. The second rope has a tension of =672.1N

The third rope has a tension of =382.78N

Anticlockwise torque is considered good, whereas clockwise torque is considered negative.

See the step by step solution

## Step 1: Conceptualize:

Consider this: For any item to be in equilibrium, the net forces operating on it must be zero, as well as the net torque acting on it.

## Step 2: Categorize:

Because the plank must be balanced,

The sum of upward and downward forces equals the sum of downward forces.

$\text{i.e.}\Sigma {F}_{\text{up}}=\Sigma {F}_{\text{dimn}}$

The torques around a point should add up to zero.

$\begin{array}{l}\text{i.e.}\Sigma \tau =0\\ \end{array}$

## Step 3: Analyze: The net vertical force is (from the free body diagram)

$\begin{array}{l}{T}_{2}+{T}_{1}\mathrm{sin}{40}^{°}=Mg+w\\ {T}_{2}+{T}_{1}\mathrm{sin}{40}^{°}=\left(30\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{2}\right)+700\text{N}\\ {T}_{2}+{T}_{1}\mathrm{sin}{40}^{°}=994\text{N}\end{array}$

The horizontal force is net.

$-{T}_{3}={T}_{1}\mathrm{cos}{40}^{°}$

In a counterclockwise route, take torque around the location where the man stood.

$\begin{array}{l}-{T}_{2}d-\mathrm{Mg}\left(\frac{L}{2}-d\right)+{T}_{1}\mathrm{sin}{40}^{°}\left(L-d\right)=0\\ -{T}_{2}\left(0.5\text{m}\right)-147\text{N}+{T}_{1}\mathrm{sin}{40}^{°}\left(2\text{m}-0.5\text{m}\right)=0\\ {T}_{2}=3{T}_{1}\mathrm{sin}{40}^{°}-294\end{array}$

## Step 4: Determining the tension:

Using the equations above,

$\begin{array}{l}3{T}_{1}\mathrm{sin}{40}^{°}-294+{T}_{1}\mathrm{sin}{40}^{°}=994\text{N}\\ {T}_{1}\mathrm{sin}{40}^{°}=322\text{N}\\ {T}_{1}=500.94\text{N}\\ {T}_{1}=501\text{N}\end{array}$

The second rope has a tension of

$\begin{array}{l}{T}_{2}=3{T}_{1}\mathrm{sin}{40}^{°}-294\\ =3\left(501\right)\mathrm{sin}{40}^{°}-294\\ =672.1\text{N}\end{array}$

The third rope has a tension of

$\begin{array}{l}{T}_{3}=-{T}_{1}\mathrm{cos}{40}^{°}\\ =-\left(501\text{N}\right)\mathrm{cos}{40}^{°}\\ =383.78\text{N}\end{array}$

## Step 5: Conclusion:

Anticlockwise torque is considered good, whereas clockwise torque is considered negative. ### Want to see more solutions like these? 