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 Orin and Anita must exert to keep the man horizontal, we can use the principle of torque equilibrium. Torque is the product of force and the distance from the point of rotation.
In this case, the point of rotation or pivot is the center of balance of the 60kg man, which is located 1m from Anita. For the man to remain horizontal (in rotational equilibrium), the total torques acting on the man must be equal and opposite.
Let's break down the calculation step by step:
1. Calculate the torques exerted by the man's weight and the force exerted by Orin and Anita. Torque is calculated using the equation:
Torque = Force x Distance
The torque exerted by the man's weight is given by:
Torque_man = Weight_man x Distance_to_center_of_balance
The torque exerted by Orin and Anita is given by:
Torque_stretcher = Force_stretcher x Distance_to_center_of_balance
2. Set up the equation for the torque equilibrium:
Torque_man = Torque_stretcher
Weight_man x Distance_to_center_of_balance = Force_stretcher x Distance_to_center_of_balance
3. Substitute the given values into the equation. The weight of the man is 60kg multiplied by the acceleration due to gravity (9.8 m/s²):
60kg x 9.8 m/s² x 1m = Force_stretcher x 1m
4. Solve for the force exerted by both Orin and Anita (Force_stretcher):
Force_stretcher = (60kg x 9.8 m/s² x 1m) / 1m
Force_stretcher = 588 Newtons
Therefore, Orin and Anita must exert a total force of 588 Newtons to keep the man horizontal.