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.
Does a figure go with this question?
I have no clue what you are talking about.
To calculate the normal force exerted by the floor on each hand, we need to consider the forces acting on the person in this position.
In this case, we have two forces acting on the person: the weight of the person and the force exerted by each hand.
1. Weight of the person:
The weight of the person can be calculated using the formula:
Weight = mass × gravity
Assuming a standard gravity of 9.8 m/s^2, we can calculate the weight of the person.
2. Force exerted by each hand:
Since the person is holding this position, we assume that the forces on both hands are the same. Let's call this force F.
Now, let's calculate the normal force exerted by the floor on each hand.
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force exerted by the floor on each hand is equal to the weight of the person plus the force exerted by each hand, since the person is at rest. Mathematically, we can express this as:
Normal force = Weight + 2 × F (since there are two hands)
Substituting the weight of the person and the force exerted by each hand:
Normal force = Weight + 2 × F
= (mass × gravity) + 2 × F
Now, we can calculate the normal force using the given values:
L1 = 0.860 m
L2 = 0.437 m
Assuming that the person is at rest in this position, L1 and L2 represent the distances from the two hands to the center of mass of the person.
Since the person is at rest, we can assume that the sum of the torques caused by the weight and the forces on each hand is equal to zero.
The torque caused by the weight is given by:
Torque_weight = (Weight × L1) + (Weight × L2)
The torque caused by the forces on each hand is given by:
Torque_force = (F × L1) + (F × L2)
Setting these two torques equal to each other:
(Weight × L1) + (Weight × L2) = (F × L1) + (F × L2)
Now, we can solve for F:
Weight × L1 + Weight × L2 - F × L1 - F × L2 = 0
We rearrange the equation to isolate F:
F × L1 + F × L2 = Weight × L1 + Weight × L2
F (L1 + L2) = Weight (L1 + L2)
F = Weight
Therefore, the force exerted by each hand (F) is equal to the weight of the person.
To get the normal force, we substitute the weight into the formula:
Normal force = Weight + 2 × F
Normal force = Weight + 2 × Weight
Normal force = 3 × Weight
So, the normal force exerted by the floor on each hand is 3 times the weight of the person.