How to lower from the roof of height $h = 16 m$ a weight of mass $m = 45 kg$ with the help of a rope, whose breaking resistance $T$ is $400 N$? The velocity of the body at the moment of impact with the ground must not exceed $v_{max} = 7 m/s$. The length of the rope is slightly greater than the height of the house.
Decision:
The force of gravity of a load $mg = 441 N$ exceeds the tensile strength of the string $T = 400 N$, and in free fall the load reaches the ground with a velocity $v =\sqrt {2gh} = 18 m/s$,
increasing $ v_{max} = 7 m/s$.
If a weight tied to a rope is not lowered with constant velocity, but with some acceleration $a$, the force of tension of the thread $F$ will be less than the force of gravity td. Applying Newton's second law to the motion of the load, we have:
$mg - F = ma$
or
$F=m(g-a)$
The fulfillment of the requirement $F < T$ leads to the inequality
$m(g-a) < T $.
This inequality is equivalent to the following:
$a > g - \frac{T}{m} = 0.9 m/s^{2}$.
So, if the load is lowered with acceleration $a > 0.9 m/s^{2}$, the tension force of the rope will not exceed $T = 400 N$. In this case, the velocity of the load at the moment of impact with the ground will not exceed the value of
$v = \sqrt{2ah} = 5.4 m/s$,
which is less than $v_{max} = 7 m/s$
Decision:
The force of gravity of a load $mg = 441 N$ exceeds the tensile strength of the string $T = 400 N$, and in free fall the load reaches the ground with a velocity $v =\sqrt {2gh} = 18 m/s$,
increasing $ v_{max} = 7 m/s$.
If a weight tied to a rope is not lowered with constant velocity, but with some acceleration $a$, the force of tension of the thread $F$ will be less than the force of gravity td. Applying Newton's second law to the motion of the load, we have:
$mg - F = ma$
or
$F=m(g-a)$
The fulfillment of the requirement $F < T$ leads to the inequality
$m(g-a) < T $.
This inequality is equivalent to the following:
$a > g - \frac{T}{m} = 0.9 m/s^{2}$.
So, if the load is lowered with acceleration $a > 0.9 m/s^{2}$, the tension force of the rope will not exceed $T = 400 N$. In this case, the velocity of the load at the moment of impact with the ground will not exceed the value of
$v = \sqrt{2ah} = 5.4 m/s$,
which is less than $v_{max} = 7 m/s$