The drawing shows a person (weight W = 586 N, L1 = 0.837 m, L2 = 0.397 m) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position.
force on each hand N
force on each foot N
To find the normal force exerted by the floor on each hand and each foot, we can use the principle of equilibrium. In this case, since the person holds the position, the sum of the forces in the vertical direction (upwards and downwards) must be zero.
Let's analyze each force separately:
1. Force on each hand:
The person's weight is acting downwards, and the normal force from the floor is acting upwards. Let's assume the person's hands are spread apart evenly, so the weight is evenly distributed between the two hands. Therefore, the weight is divided by 2.
Weight of the person (W) = 586 N
Force on each hand = Weight / 2
= 586 N / 2
= 293 N
Therefore, the normal force exerted by the floor on each hand is 293 N.
2. Force on each foot:
Similar to the hands, the person's weight is acting downwards, and the normal force from the floor is acting upwards. Let's assume the person's feet are spread apart evenly, so the weight is evenly distributed between the two feet. Therefore, the weight is divided by 2.
Weight of the person (W) = 586 N
Force on each foot = Weight / 2
= 586 N / 2
= 293 N
Therefore, the normal force exerted by the floor on each foot is also 293 N.
To find the normal force exerted by the floor on each hand and each foot, we need to use the concept of torque balance.
In this case, the person is doing push-ups, which means their body is in equilibrium. This means that the clockwise and counterclockwise torques acting on the person's body should balance each other out.
Let's start with the counterclockwise torque due to the weight of the person. The weight can be represented by a force acting downwards at the center of mass of the person, which is equal to W = 586 N. The distance from the center of mass to the hands is L1 = 0.837 m, and the distance to the feet is L2 = 0.397 m.
The counterclockwise torque due to the weight of the person about the hands can be calculated by multiplying the weight by the distance to the hands:
Torque_hands = W * L1
Similarly, the counterclockwise torque due to the weight of the person about the feet can be calculated by multiplying the weight by the distance to the feet:
Torque_feet = W * L2
In order for the person to be in equilibrium, the clockwise torques exerted by the floor on the hands and feet must balance the counterclockwise torques due to the weight.
The clockwise torque exerted by the floor on the hands is equal to the normal force on each hand multiplied by the distance to the hands, which is L1. Similarly, the clockwise torque exerted by the floor on the feet is equal to the normal force on each foot multiplied by the distance to the feet, which is L2.
Setting up the torque balance equations, we have:
Clockwise torque_hands = Counterclockwise torque_hands
Normal force_hands * L1 = W * L1
Clockwise torque_feet = Counterclockwise torque_feet
Normal force_feet * L2 = W * L2
Now, we can solve these equations to find the normal forces on each hand and each foot.
Normal force_hands = W
Normal force_feet = W
Therefore, the normal force exerted by the floor on each hand and each foot is equal to the weight of the person, which is 586 N.