Orin and Anita rush a 60kg man from the scene of an accidnet, carrying him on a uniform 3kg stretcher held by the ends. The stretcher is 2.6m long and the man's center of balance is 1m from Anita. How much force must they both exert to keep him horizontal?
To determine the amount of force required to keep the man horizontal, we can use the principle of moments or torque. The principle of moments states that the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
In this scenario, we have two forces at play: the weight of the man and the force exerted by Orin and Anita. Let's denote the force exerted by Orin as F_orin and the force exerted by Anita as F_anita.
The weight of the man can be calculated using the formula:
Weight = mass * gravity
Given that the mass of the man is 60kg and the acceleration due to gravity is approximately 9.8 m/s², we have:
Weight = 60kg * 9.8 m/s² = 588 N
Now, let's consider the moments. The moment produced by the weight of the man can be found by multiplying the weight by the distance of the man's center of balance from Anita, which is 1m in this case:
Moment_man = Weight * Distance = 588 N * 1m = 588 N·m
Since the stretcher is held at both ends, the total length of the stretcher acts as the lever arm for the force exerted by Orin and Anita. The total length of the stretcher is 2.6m.
To balance the horizontal position of the man, the sum of the moments must be zero. So, we have:
Clockwise Moment = Anticlockwise Moment
F_orin * 2.6m = Moment_man
F_orin = Moment_man / 2.6m
F_orin = 588 N·m / 2.6m ≈ 226.15 N
Therefore, both Orin and Anita must exert approximately 226.15 Newtons of force each to keep the man horizontal.