Question

Please please helpjejejnebwbww​

Answer 1

Answer:

C. 540 N

Explanation:

Let suppose that system formed by the athlete and the load are in equilibrium. By Newton's Laws of Motion, we use the moment equation with respect to the feet of the athlete to determine the upward force exerted by his two arms:

$\Sigma M = -F\cdot r_{1} + m_{L}\cdot g \cdot r_{2} + m_{A}\cdot g \cdot r_{3} = 0$

$F = \frac{(m_{L}\cdot r_{2} + m_{A}\cdot r_{3})\cdot g}{r_{1}}$ (1)

Where:

$m_{L}$ - Mass of the load, in kilograms.

$m_{A}$ - Mass of the athlete, in kilograms.

$g$ - Gravitational acceleration, in meters per square second.

$r_{1}$, $r_{2}$, $r_{3}$ - Distances with respect to the feet, in meters.

$F$ - Upward force exerted by his two arms, in newtons.

If $m_{L} = 6\,kg$, $r_{2} = 1.20\,m$, $m_{A} = 70\,kg$, $r_{3} = 0.90\,m$, $g = 9.807\,\frac{m}{s^{2}}$ and $r_{1} = 1.30\,m$, then the upward force is:

$F = \frac{[(6\,kg)\cdot (1.20\,m)+(70\,kg)\cdot (0.90\,m)]\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{1.30\,m}$

$F = 529.578\,N$

The upward force exerted by his two arms is 529.578 newtons. (Right answer: C)