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Q25P

Expert-verifiedFound in: Page 382

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 ${\mathit{d}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{500}}{\mathit{m}}$**** 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.

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.

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.

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

The net vertical force is (from the free body diagram)

$\begin{array}{l}{T}_{2}+{T}_{1}\mathrm{sin}{40}^{\xb0}=Mg+w\\ {T}_{2}+{T}_{1}\mathrm{sin}{40}^{\xb0}=(30\text{kg})(9.8\text{m}/{\text{s}}^{2})+700\text{N}\\ {T}_{2}+{T}_{1}\mathrm{sin}{40}^{\xb0}=994\text{N}\end{array}$

The horizontal force is net.

$-{T}_{3}={T}_{1}\mathrm{cos}{40}^{\xb0}$

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

$\begin{array}{l}-{T}_{2}d-\mathrm{Mg}(\frac{L}{2}-d)+{T}_{1}\mathrm{sin}{40}^{\xb0}(L-d)=0\\ -{T}_{2}(0.5\text{m})-147\text{N}+{T}_{1}\mathrm{sin}{40}^{\xb0}(2\text{m}-0.5\text{m})=0\\ {T}_{2}=3{T}_{1}\mathrm{sin}{40}^{\xb0}-294\end{array}$

Using the equations above,

$\begin{array}{l}3{T}_{1}\mathrm{sin}{40}^{\xb0}-294+{T}_{1}\mathrm{sin}{40}^{\xb0}=994\text{N}\\ {T}_{1}\mathrm{sin}{40}^{\xb0}=322\text{N}\\ {T}_{1}=500.94\text{N}\\ {T}_{1}=501\text{N}\end{array}$

The second rope has a tension of

$\begin{array}{l}{T}_{2}=3{T}_{1}\mathrm{sin}{40}^{\xb0}-294\\ =3\left(501\right)\mathrm{sin}{40}^{\xb0}-294\\ =672.1\text{N}\end{array}$

The third rope has a tension of

$\begin{array}{l}{T}_{3}=-{T}_{1}\mathrm{cos}{40}^{\xb0}\\ =-(501\text{N})\mathrm{cos}{40}^{\xb0}\\ =383.78\text{N}\end{array}$

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

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