Torques and Tug-of-War. In a study of the biomechanics of tug-of-war, atall, competitor in the middle of the line is considered to be a rigid body leaning back at an angle of $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$to the vertical. The competitor is pulling on a rope that is held horizontal a distance 1.50m 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 ${\mathit{T}}_{\mathbf{1}}\mathbf{=}\mathbf{1160}\mathit{N}$. 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 . 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 again leaning back at 30.$\mathbf{0}\mathbf{°}$to the vertical, but now the height at which the rope is held above but still parallel to the ground is varied. The tension in the rope in front of the competitoris measured as a function of the shortest distance between the rope and the ground (the holding height). Tension${\mathit{T}}_{\mathbf{1}}$is found to decrease as the holding height increases. What could explain this observation? As the holding height increases,

1. the moment arm of the rope about his feet decreases due to the angle that his body makes with the vertical;
2. the moment arm of the weight about his feet decreases due to the angle that his body makes with the vertical;
3. a smaller tension in the rope is needed to produce a torque sufficient to balance the torque of the weight about his feet;
4. his center of mass moves down to compensate, so less tension in the rope is required to maintain equilibrium

Step by step solution

01

Identification of given data

Here we have,

$$\begin{gathered} {\mathrm{T}}_1=1160\mathrm{N} \\ {\mathrm{\mu }}_{\mathrm{s}}=0.65 \\ \mathrm{\theta }=30° \end{gathered}$$

02

Weight

Weight is the force of gravity acting on an object.

W = m. g

Where is weight of object, m is mass of object and g is gravitational force.

03

Finding the true option from the following options

Free body diagram is given by,

Let the distance from feet to rope is X.

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

$\begin{array}{r}{T}_2\times \cos \theta +W\left(1.00\mathrm{m}\right)\sin \theta -T_1\times \cos \theta =0\\ x\left(T_1-T_2\right)=\frac{\left(mg\right)\left(1.00\mathrm{m}\right)\sin \theta }{\cos \theta }\\ =\frac{\left(80\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1.00\mathrm{m}\right)\sin \theta }{\cos \theta }\\ =784\tan \theta \end{array}$

So, we can say that when X is increasing$T_1-T_2$ is decreasing.

Also, we can see that when$T_1$ decreases, x increases because RHS of above equation is constant.

Hence, option (c) is right explanation.

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