A man doing a slow push-up is approximately in static equilibrium. His body is horizontal, with his weight of 755 N supported by his hands and feet, which are 1.34 m apart. One hand rests on a spring scale, which reads 270 N. If each hand bears an equal weight, how far from the shoulders is the man's center of gravity?
To find the distance from the man's center of gravity to his shoulders, we need to understand the concept of center of gravity and how it relates to static equilibrium.
The center of gravity (COG) is the point at which the weight of an object can be considered to act. It is usually located near the geometric center of an object. When an object is in static equilibrium, the sum of the forces and torques acting on it is zero.
In this case, we have a man doing a slow push-up. The weight of his body is 755 N, which is supported by his hands and feet. Since we are assuming that each hand bears an equal weight, each hand must support half the total weight of his body, which is 377.5 N.
The hand resting on the spring scale reads 270 N. This means that the other hand supports the remaining weight of 377.5 N minus 270 N, which is 107.5 N.
Since the man is in static equilibrium, the sum of the forces acting vertically must be zero. This implies that the vertical components of the forces acting on the man's body must balance each other out.
The weight supported by each hand (270 N and 107.5 N) is equal to the vertical component of the man's weight. So, by using a proportion, we can determine the distance between the hand on the spring scale and the man's center of gravity.
We can set up the following proportion:
270 N / 377.5 N = x / 1.34 m
Simplifying this equation, we get:
x = (270 N / 377.5 N) * 1.34 m
x ≈ 0.96 m
Therefore, the man's center of gravity is approximately 0.96 meters away from his shoulders.