Assume L1 = 0.860 m and L2 = 0.437 m. Calculate the normal force exerted by the floor on each hand, assuming that the person holds this position. Assume the force on the left hand is the same as the force on the right hand.
To calculate the normal force exerted by the floor on each hand, we can use the principle of moments or torques.
When a person holds this position, we can assume that their body is in equilibrium, meaning that the sum of the torques acting on their body is zero.
In this case, we have two torques acting on the person: one due to the weight of their body acting at the center of mass, and the other due to the normal forces at each hand acting at a distance from the center of mass.
The torque due to the weight can be calculated as the weight (mg) multiplied by the distance from the center of mass (L1) to the point of contact with the floor (the left hand and the right hand).
Similarly, the torque due to the normal forces at each hand can be calculated as the normal force multiplied by the distance from the center of mass (L2) to the point of contact with the floor.
Since we assume the forces on the left and right hand are equal, we can set up the equation:
Weight * L1 = 2 * Normal Force * L2
We know the values for L1 and L2, so we can solve the equation for the normal force.
Substituting the values, we have:
Normal Force = (Weight * L1) / (2 * L2)
To calculate the weight, we can use the formula:
Weight = mass * gravitational acceleration
Since we don't have the mass of the person, we can assume a standard value for gravitational acceleration, which is approximately 9.8 m/s².
Substituting the values, we have:
Weight = mass * 9.8 m/s²
Finally, we can substitute this into the equation for the normal force to get the final answer.