A man doing push-ups pauses. His mass is 80-kg. Determine the normal force exerted by the floor on each hand and on each foot.
a = 40 cm
b = 95 cm
c = 30 cm
W = mg = 784 N
You have to know the center of gravity. I do not know what a,b, c mean above.
Sum moments around any point. Sum vertical forces. Those two equations should do it.
To determine the normal force exerted by the floor on each hand and foot of the man doing push-ups, we need to consider the forces acting on each body part.
1. Determine the weight of the man:
The weight of an object is given by the formula W = mg, where m is the mass and g is the acceleration due to gravity. In this case, the mass (m) is given as 80 kg. Therefore, the weight (W) of the man is calculated as follows:
W = 80 kg × 9.8 m/s^2 = 784 N
2. Divide the weight between the hands and feet:
Since the man is doing push-ups, his weight is distributed between his hands and feet. To calculate the normal forces on each hand and foot, we need to consider the geometry of the push-up position.
- For the hands:
Let's assign point A as the contact point of one hand with the floor. The normal force exerted by the floor on the hand at this point is equal to the weight of the man, since there are no other vertical forces acting on the hand in this position. Therefore, the normal force on one hand is 784 N.
- For the feet:
Let's assign point B as the contact point of one foot with the floor. In the push-up position, the feet are typically closer together than the hands, forming a triangle. To calculate the normal force on one foot, we need to resolve the weight vector into two components.
First, calculate the horizontal distance between the hands using the given dimensions a and b:
d = b - a = 95 cm - 40 cm = 55 cm = 0.55 m
Next, calculate the vertical distance between the hands and the feet. In this case, it is given as c = 30 cm = 0.30 m.
Using these dimensions, we can calculate the weight of the man supported by one foot using trigonometry. The vertical component of the weight is given by W * (c / d), and the horizontal component is given by W * (a / d).
- Vertical component:
Weight supported by one foot = W * (c / d) = 784 N * (0.30 m / 0.55 m) = 430.91 N
- Horizontal component:
Weight supported by hands = W * (a / d) = 784 N * (0.55 m / 0.55 m) = 784 N
Therefore, the normal forces exerted by the floor on each foot and on each hand are as follows:
- Normal force on each hand: 784 N
- Normal force on each foot: Vertical component - 430.91 N, Horizontal component - 784 N