'n Man beland in die moeilijkheid terwyl hy in 'n dam swem - NSC Physical Sciences - Question 2 - 2022 - Paper 1

The free body diagram should include the forces acting on the man-bin combination:

1. Weight (downward): $F_g = m imes g$
2. Tension in the rope ($T$): acting upward along the direction of the rope.
3. Applied force (horizontal): the wrywingkrag (300 N) acting against the motion.

We can use Newton's second law to find the tension in the rope. By applying the equation:

$F_{net} = ma$,

we know the net force is the difference between the upward force (tension) and the downward force (weight) applied to the man-bin combination:

1. Calculate the weight:
   • $F_g = (70 + 4) kg imes 9.8 m/s^2 = 730.8 N$
2. The net force acting:
   • $F_{net} = F_{T} - F_{g} = 300 N$ (as it's opposing the wrywingkrag)
3. Rearranging the equation will give us:
   • $F_T = F_{g} + F_{net} = 730.8 + 300 = 1030.8 N$

Hence, the tension in the rope is 1030.8 N.

If the helicopter accelerates while pulling the man, the tension in the rope would increase because the additional force needed to accelerate the man-bin combination would have to be accounted for. Thus, we would have:

$F_{net} = T - F_g - F_f,$

where $F_f$ is the frictional force. The net force equation would need to factor in acceleration, thereby making:

$T = F_g + F_f + ma,$

This means an increase in tension in the rope due to the added acceleration.