A man doing a slow push-up is approximately in static equilibrium. His body is horizontal, with his weight of 750 N supported by his hands and feet, which are 1.36 m apart. One hand rests on a spring scale, which reads 268 N. If each hand bears an equal weight, how far from the shoulders is the man's center of gravity?

Question

A man doing a slow push-up is approximately in static equilibrium. His body is horizontal, with his weight of 750 N supported by his hands and feet, which are 1.36 m apart. One hand rests on a spring scale, which reads 268 N. If each hand bears an equal weight, how far from the shoulders is the man's center of gravity?

Answer 1

I assume hands under shoulders

x = CG to feet

hands hold 268*2 = 536 N
so
536 * 1.36 = 750 x
x = .972 m from CG to feet
1.36 - .972 = .388 CG to hands (shoulders)

Answer 2

To find the distance from the man's center of gravity to his shoulders, we can use the concept of torque. Torque is the product of force and distance, and it measures the tendency of a force to rotate an object around a pivot point.

In this case, the pivot point can be considered as the location of the man's shoulders. Since the man is in static equilibrium, the torques acting on his body must balance each other out.

Here's how to calculate the distance from the man's center of gravity to his shoulders:

1. First, let's determine the torque exerted on each hand. We know that the weight of the man's body is 750 N, and the distance between his hands and feet is 1.36 m. Since we're assuming each hand bears an equal weight, the weight on each hand is 750 N / 2 = 375 N.

2. To calculate the torque on each hand, we multiply the weight on each hand by the distance between the hand and the pivot point (shoulders). Therefore, the torque on each hand is 375 N × 1.36 m = 510 N·m.

3. Since the man is in static equilibrium, the total torque on his body must be zero. The torque exerted by the hand on the spring scale is 268 N (as given in the problem). Therefore, the torque exerted by the hand on the opposite side (without the scale) must be equal in magnitude but opposite in direction.

4. Let's call the distance from the man's center of gravity to his shoulders x. This means that the distance from each hand to the center of gravity is x/2. Since the torques exerted by both hands must balance each other out, we can set up the following equation:

(375 N) × (x/2) = (268 N) × (1.36 m)

5. Simplifying this equation, we get:

(375 N) × (x/2) = 363.68 N·m

6. Cross-multiplying and solving for x, we find:

375 N × x = 2 × 363.68 N·m

x = (2 × 363.68 N·m) / 375 N

x = 1.95 m

Therefore, the man's center of gravity is located approximately 1.95 meters from his shoulders.