Question

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.

Short answer

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.

Step-by-step solution

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.

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.

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

Analyze:

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

$\begin{array}{l}{T}_2+T_1\sin {40}^{°}=Mg+w\\ T_2+T_1\sin {40}^{°}=(30\text{kg})(9.8\text{m}/{\text{s}}^2)+700\text{N}\\ T_2+T_1\sin {40}^{°}=994\text{N}\end{array}$

The horizontal force is net.

$-T_3=T_1\cos {40}^{°}$

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

$\begin{array}{l}-T_{2}d-\mathrm{Mg}(\frac{L}{2}-d)+T_1\sin {40}^{°}(L-d)=0\\ -T_2(0.5\text{m})-147\text{N}+T_1\sin {40}^{°}(2\text{m}-0.5\text{m})=0\\ T_2=3T_1\sin {40}^{°}-294\end{array}$

Determining the tension:

Using the equations above,

$\begin{array}{l}3T_1\sin {40}^{°}-294+T_1\sin {40}^{°}=994\text{N}\\ T_1\sin {40}^{°}=322\text{N}\\ T_1=500.94\text{N}\\ T_1=501\text{N}\end{array}$

The second rope has a tension of

$\begin{array}{l}{T}_2=3T_1\sin {40}^{°}-294\\ =3\left(501\right)\sin {40}^{°}-294\\ =672.1\text{N}\end{array}$

The third rope has a tension of

$\begin{array}{l}{T}_3=-T_1\cos {40}^{°}\\ =-(501\text{N})\cos {40}^{°}\\ =383.78\text{N}\end{array}$

Conclusion:

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