A man doing push-ups pauses in the position shown in the figure . His mass = 68 kg.
Determine the normal force exerted by the floor on each hand.
Determine the normal force exerted by the floor on each foot.
Well, if this man is pausing in a push-up position, I guess you could say he's really "feeling the burn!" Now, let's get down to business and find those normal forces.
First, let's focus on the normal forces on each hand. In this position, his hands are stationary and not moving, so the normal force exerted by the floor on each hand would simply be his weight, which is his mass multiplied by the acceleration due to gravity (9.8 m/s^2). So, the normal force on each hand would be 68 kg * 9.8 m/s^2.
Next, let's move on to the normal forces on each foot. Since his entire body weight is distributed between two feet, the normal force on each foot would be half of his body weight. So, the normal force on each foot would be half of his weight, which is half of 68 kg * 9.8 m/s^2.
I hope that clears things up! Remember, when it comes to push-ups, if the numbers don't add up, just try adding a few more reps. Keep pushing through!
To determine the normal force exerted by the floor on each hand and foot, we need to consider the forces acting on the man and apply Newton's third law.
1. Normal force on each hand:
Since the man is in equilibrium, the sum of the vertical forces acting on him must be zero. These forces include the weight (mg) and the normal force (N) exerted by the floor on each hand.
Considering the forces in the vertical direction:
ΣFy = 0
N - mg = 0
N = mg
So, the normal force exerted by the floor on each hand is equal to the weight, which is given by:
N = m * g
N = 68 kg * 9.8 m/s^2
N ≈ 666.4 N (rounded to the nearest whole number)
2. Normal force on each foot:
Similar to the hand, the normal force on each foot is also the weight of the man, which is given by:
N = m * g
N = 68 kg * 9.8 m/s^2
N ≈ 666.4 N (rounded to the nearest whole number)
Therefore, the normal force exerted by the floor on each hand and foot is approximately 666.4 N each.
To determine the normal force exerted by the floor on each hand and foot, we need to consider the forces acting on the man.
First, let's analyze the forces acting on the man:
1. Weight (mg): The man's weight is the force exerted by gravity, given by the product of his mass (m) and the acceleration due to gravity (g = 9.8 m/s^2). So, the weight is given by W = mg.
2. Normal force (N): The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface. In this case, the normal force refers to the force exerted by the floor on the man's hands and feet.
Now let's determine the normal force exerted by the floor on each hand:
When the man is in the push-up position, the normal force on each hand is equal to the weight of the man divided by the number of hands supporting the weight. In this case, the man's weight is given as 68 kg, and he is supporting his weight with two hands. So, the normal force on each hand is given by N_hand = W / 2.
Normal force exerted by the floor on each hand = N_hand = W / 2 = (68 kg) * (9.8 m/s^2) / 2 = 333.4 N.
Now let's determine the normal force exerted by the floor on each foot:
When the man is in the push-up position, the normal force on each foot is equal to the weight of the man divided by the number of feet supporting the weight. In this case, the man's weight is given as 68 kg, and he is supporting his weight with two feet. So, the normal force on each foot is given by N_foot = W / 2.
Normal force exerted by the floor on each foot = N_foot = W / 2 = (68 kg) * (9.8 m/s^2) / 2 = 333.4 N.
Therefore, the normal force exerted by the floor on each hand and foot is 333.4 N.