91影视

Given that the yoga exercise 鈥淒ownward-Facing Dog鈥� requires stretching your hands straight out above your head and bending down to lean against the floor. This exercise is performed by a 750-N person as shown in the fig. When he bends his body at the hip to a angle between his leg and trunk, his legs, trunk, head and arms have the dimensions indicated. Furthermore, his legs and feet weigh a total of 277 N, and their center of mass is 41 cm from his hip, measured along his legs. The person鈥檚 trunk, head and arms weigh 473 N, and their center of gravity is 65 cm from his hip, measured along the upper body.

Weight of feet and legs ${\mathrm{w}}_1=277\mathrm{N}$

Weight of trunk, ${\mathrm{w}}_t=473\mathrm{N}$

Let $n_f$and $f_f$are the total normal and friction forces exerted on his feet and $n_h$and $f_h$are those forces on his hands.

Determine the angle as:

$$\begin{gathered} 胃+蠁=90掳 \\ 胃=56.3掳 \\ 蠁=33.7掳 \end{gathered}$$

So ${\mathrm{x}}_1=\left(90\mathrm{cm}\right)\mathrm{肠辞蝉胃}=50\mathrm{cm}$

And ${\mathrm{x}}_2=\left(135\mathrm{cm}\right)\cos 蠁=112\mathrm{cm}$

Consider the formula for the torque is

$\mathbf{蟿}\mathbf{=}\mathbf{FI}$

Here, Fis force exerted and l is moment arm.

Consider the following diagram:

According to condition of equilibrium with pivot at his feet.

Net torque is zero.

So

$\begin{array}{r}{n}_h\left(162\mathrm{cm}\right)鈭�\left(277\mathrm{N}\right)\left(27.2\mathrm{cm}\right)鈭�\left(473\mathrm{N}\right)\left(103.8\mathrm{cm}\right)=0\\ n_h=350\mathrm{N}\end{array}$

Hence, there is normal force of at each hand.

Now,

So

Hence, there is normal force of $\frac{350\mathrm{N}}{2}=175\mathrm{N}$at each foot.

Now, $n_f+n_h-w_1-w_t=0$

So

$\begin{array}{r}{n}_f=w_1+w_t鈭�n_h\\ =750\mathrm{N}鈭�350\mathrm{N}\\ =400\mathrm{N}\end{array}$

Hence, there is normal force of$\frac{400\mathrm{N}}{2}=200\mathrm{N}$ at each foot.

Consider the formula for the free body diagram.

Applying the condition of equilibrium with pivot at his hip.

Net torque is zero.

So $f_f\left(74.9\mathrm{cm}\right)+w_1\left(22.8\mathrm{cm}\right)鈭�n_f\left(50\mathrm{cm}\right)=0$

On solving $f_f=\frac{\left(400\mathrm{N}\right)\left(50\mathrm{cm}\right)鈭�\left(277\mathrm{N}\right)\left(22.8\mathrm{cm}\right)}{74.9\mathrm{cm}}=182.7\mathrm{N}$

So there is frictional force of 91 N on each feet.

Also frictional force on feet and hands are same.

So frictional force on each hand is 91 N.