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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 ${\mathit{T}}_{\mathbf{1}}\mathbf{=}\mathbf{1160}\mathit{N}$. 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.

If he leans slightly farther back (increasing the angle between his body and the vertical) but remains stationary in this new position, which of the following statements is true? Assume that the rope remains horizontal.

1. The difference between${\mathit{T}}_{\mathbf{1}}$and${\mathit{T}}_{\mathbf{2}}$will increase, balancing the increased torque about his feet that his weight produces when he leans farther back;
2. The difference betweenandwill decrease, balancing the increased torque about his feet that his weight produces when he leans farther back;
3. neither${\mathit{T}}_{\mathbf{1}}$nor${\mathit{T}}_{\mathbf{2}}$will change, because no other forces are changing;
4. Both${\mathit{T}}_{\mathbf{1}}$and${\mathit{T}}_{\mathbf{2}}$will change, but the difference between them will remain the same.

Step by step solution

01

Identification of given data

Here we have, m = 80 kg

$$\begin{gathered} T_1=1160N \\ {渭}_s=0.65 \\ 胃=30掳 \end{gathered}$$

02

Tension

A force throughout a medium's length is known as tension, particularly a force carried by a flexible medium like a rope or cable.

03

Finding the true option from the following options

Free body diagram is given by,

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

$\underset{}{鈭�}{蟿}_A=0$

$\begin{array}{r}{T}_2\left(1.50\mathrm{m}\right)\cos 鈦�胃+W\left(1.00\mathrm{m}\right)\sin 鈦�胃鈭�T_1\left(1.50\mathrm{m}\right)\cos 鈦�胃=0\\ T_1鈭�T_2=\frac{\left(mg\right)\left(1.00\mathrm{m}\right)\sin 鈦�胃}{\left(1.50\mathrm{m}\right)\cos 鈦�胃}\\ =\frac{\left(80\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1.00\mathrm{m}\right)\sin 鈦�胃}{\left(1.50\mathrm{m}\right)\cos 鈦�胃}\\ =522.66\tan 鈦�胃\end{array}$

Here if$胃$increases,$\tan 胃$increases.

Then, $T_1-T_2$increases. Here, $T_1-T_2$depends on the value of $\tan 胃$

Hence, option (a) is correct.

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