Torques and Tug-of-War. In a study of the biomechanics of tug-of-war, a 2.0 mtall, 80.0 kgcompetitor in the middle of the line is considered to be a rigid body leaning back at an angle of $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$to the vertical. The competitor is pulling on a rope that is held horizontal a distance 1.50mfrom 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 ${\mathit{T}}_{\mathbf{1}}\mathbf{=}\mathbf{1160}\mathit{N}$. Since there is friction between the rope and his hands, the tension in the rope behind him, ${\mathit{T}}_{\mathbf{2}}$, is not equal to ${\mathit{T}}_{\mathbf{1}}$. 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.

His body is leaning back at $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$to the vertical, but the coefficient of static friction between his feet and the ground is suddenly reduced to 0.50. What will happen?

1. His entire body will accelerate forward;
2. his feet will slip forward;
3. his feet will slip backward;
4. his feet will not slip.

Step by step solution

01

Identification of given data

Here we have, m = 80 kg

$$\begin{gathered} T_1=1160N \\ {\mu }_s=0.65 \\ \theta =30° \end{gathered}$$

02

Friction force

The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction.

${\mathit{f}}_{\mathbf{s}}\mathbf{=}\mathit{\mu }\mathbf{·}\mathit{N}$ (1)

Where, is friction force, is constant of friction and is normal force

03

Finding what will happen if his body is leaning back at 30.0° to the vertical, but the coefficient of static friction between his feet and the ground is suddenly reduced to 0.50.

Free body diagram is given by,

Now, from free body diagram and by second condition of equilibrium we have,

$$\begin{gathered} \sum _{}{\tau }_A=0 \\ {T}_2\left(1.50m\right)\cos 30°+W\left(1.00m\right)\sin 30°-T_1\left(1.50m\right)\cos 30°=0 \\ {T}_2=\frac{T_1\left(1.50m\right)\cos 30°-\left(80kg\right)\left(9.8m/s^2\right)\left(1.00m\right)\sin 30°}{\left(1.50m\right)\cos 30} \\ =\frac{\left(1160N\right)\left(1.50m\right)\cos 30°-\left(80kg\right)\left(9.8m/s^2\right)\left(1.00m\right)\sin 30°}{\left(1.50m\right)\cos 30} \\ \approx 858.2N \end{gathered}$$

Now, from first condition of static equilibrium

$$\begin{gathered} \sum _{}{F}_x=0 \\ N-W=0 \\ N=W \\ N=\left(80kg\right)\left(9.8m/s^2\right) \\ N=784N \end{gathered}$$

Also,

$$\begin{gathered} \sum _{}{F}_x=0 \\ {T}_1-T_2-f_s=0 \\ {f}_s=T_1-T_2 \\ =1160N-858.2N \\ {f}_s=301.8N \end{gathered}$$

From equation (1) that his feet will not slip if

$$\begin{gathered} f_s\le {\mu }_{s}N \\ {f}_s\le \left(0.65\right)\left(784N\right) \\ =392N \end{gathered}$$

Also,$301.8\mathrm{N}\le 392\mathrm{N}$

So, the feet will not slip.

Hence, option (d) is correct.

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