91影视

Given that a 72.0-kg weightlifter doing arm raises holds a 7.50-kg weight. Her arm pivots around the elbow joint, starting $40.0掳$below the horizontal. Biometric measurements have shown that, together, the forearms and the hands account for 6.00% of a person鈥檚 weight. Since the upper arm is held vertically, the biceps muscle always acts vertically and is attached to the bones of the forearm 5.50 cm from the elbow joint. The center of mass of this person鈥檚 forearm-hand combination is 16.0 cm from the elbow joint, along the bones of the forearm, and she holds the weight 38.0 cm from her elbow joint.

Mass of weightlifter, $m_w=72\mathbf{kg}$

Mass held in each forarm $m_{arm}=\frac{1}{2}\left(0.06\right)\left(72\mathbf{kg}\right)=2.16\mathbf{kg}$

The weight held in each arm, w = mg where m = 7.50kg

Let T is the force exerted by biceps on forearms

E is the force exerted by the elbow and has vertical and horizontal components $E_v$and $E_h$.

Torque $蟿=FI$

Where Fis force exerted and l is distance.

The free body diagram is as follows:

Applying condition of equilibrium,

Net torque is zero.

$\underset{}{鈭�}蟿=0$

This gives,

$$\begin{gathered} \mathrm{T}\left(5.5\mathrm{cm}\right)\left(\mathrm{肠辞蝉胃}\right)-{\mathrm{m}}_{\mathrm{arm}}\left(16\mathrm{cm}\right)\left(\mathrm{肠辞蝉胃}\right)=0 \\ \mathrm{T}=\frac{16{\mathrm{m}}_{\mathrm{arm}}\mathrm{g}+38\mathrm{w}}{5.5\mathbf{cm}} \\ \mathrm{T}=\frac{16\left(2.16\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)+38\left(7.5\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{5.5\mathrm{cm}} \\ \mathrm{T}=569\mathbf{N} \end{gathered}$$

Hence biceps exerts a force of 569 N on forearms

Components of net force are zero as the arm is at rest.

This implies$\underset{}{鈭�}{F}_x$ and$\underset{}{鈭�}{F}_y$ are zero.

$\underset{}{鈭�}{F}_x=0$implies $E_h=0$

$\underset{}{鈭�}{F}_y=0$implies $T-E_v-W_{arm}-W=0$

This implies

$$\begin{gathered} E_v=T-w_{arm}-w \\ =\left(569\mathbf{N}\right)-\left(2.16\mathbf{kg}\right)\left(9.8\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\right)-\left(7.5\mathbf{kg}\right)\left(9.8\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\right) \\ =474\mathbf{N} \end{gathered}$$

Hence elbow exerts a force of 474 N on forearms in downwards direction

The weight and the pull of the biceps are both always vertical in this situation, so the factor divides out of the$\underset{}{鈭�}蟿=0$ equation in part (b). Therefore, the force stays the same on raising the arm.