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Torques and Tug-of-War. In a study of the biomechanics of tug-of-war, a 2.0 mtall, 80.0 kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{掳}$to the vertical. The competitor is pulling on a rope that is held horizontal a distance 1.50 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is ${\mathbf{T}}_{\mathbf{1}}\mathbf{=}\mathbf{1160}\mathbf{N}$. Since there is friction between the rope and his hands, the tension in the rope behind him, ${\mathbf{T}}_{\mathbf{2}}$, is not equal to ${\mathbf{T}}_{\mathbf{1}}$. His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

What is tension${\mathit{T}}_{\mathbf{2}}$in the rope behind him?

1. 590N
2. 650N
3. 860N
4. 1100N

Step by step solution

01

Identification of given data

Here we have, m = 80 kg

$T_1=1160N$

${渭}_s=0.65$

$胃=30掳$

02

Second condition of equilibrium

Which is says that,the object must also be subject to zero net torque.

$\underset{\mathbf{}}{\mathbf{鈭�}}\stackrel{\mathbf{鈫�}}{\mathbf{蟿}}\mathbf{=}\mathbf{0}$

Where$\stackrel{\mathbf{鈫�}}{\mathbf{蟿}}$ is net external torque.

03

Finding the tension T2 in the rope behind him.

Free body diagram is given by,

Now, from free body diagram and equation (1) we have,

$\underset{}{鈭�}{蟿}_A=0$

$\begin{array}{r}{T}_2\left(1.50\mathrm{m}\right)\cos 鈦�{30}^{鈭�}+W\left(1.00\mathrm{m}\right)\sin 鈦�{30}^{鈭�}鈭�T_1\left(1.50\mathrm{m}\right)\cos 鈦�{30}^{鈭�}=0\\ T_2=\frac{T_1\left(1.50\mathrm{m}\right)\cos 鈦�{30}^{鈭�}鈭�\left(mg\right)\left(1.00\mathrm{m}\right)\sin 鈦�{30}^{鈭�}}{\left(1.50\mathrm{m}\right)\cos 鈦�30}\\ =\frac{\left(1160\mathrm{N}\right)\left(1.50\mathrm{m}\right)\cos 鈦�{30}^{鈭�}鈭�\left(80\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1.00\mathrm{m}\right)\sin 鈦�{30}^{鈭�}}{\left(1.50\mathrm{m}\right)\cos 鈦�30}\\ 鈮�860\mathrm{N}\end{array}$

Therefore, the tension${\mathrm{T}}_2=860\mathrm{N}$

Hence, option (c) is correct.

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