A man doing push-ups pauses in the position shown in the figure. His mass is 75-kg. Determine the normal force exerted by the floor on each hand and on each foot.
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a = 40 cm
b = 95 cm
c = 30 cm
W = mg = 735 N
I can't figure out how to find the amount of force being exerted at the hands/feet if you don't know the mass at each location. Help would be very much appreciated.
I thought W was the center of mass?
Also, are you saying to use moment of inertia I to solve, and if so, how would I set up the equations? I'm not good at equilibrium problems.
You don't need to know the distinct masses of hands and feet... just the total mass and that it acts as if it were at the Center of Mass.
The height of the CM above the floor, although shown in the figure, will not matter in this problem.
Assume the forces on each foot are equal, and the forces on each hand are equal. The total on all four appendages must equal 735 N.
Set the total moment about a line through the feet/floor contact points equal to zero. This will let you solve for the sum of the forces on the hands. You will need to know where the center of mass of the body is. This is shown as "b" in your figure
W is the center of mass. The moment of inertia does not play a role in statics problems.
I still don't know get to set up how to solve for the forces of the hands/feet if you don't know their distinct masses and only have the center of mass...
To determine the normal force exerted by the floor on each hand and on each foot, we can make use of the principles of static equilibrium. When the man is in the paused position during a push-up, the forces acting on him are in equilibrium, which means that the sum of all the forces must be equal to zero.
Let's analyze the forces acting on the man:
1. Weight (W): The weight of the man is given as 735 N. This force acts vertically downward through the center of mass.
2. Normal force at hand (N1): This force is exerted by the floor on each hand to counteract the weight of the man. Since the hands are in contact with the floor beneath them, the normal force acts vertically upward.
3. Normal force at foot (N2): This force is exerted by the floor on each foot to counteract the weight of the man. Since the feet are in contact with the floor beneath them, the normal force acts vertically upward.
By considering the equilibrium condition, we can write the following equations:
∑Fx = 0: There is no horizontal acceleration, so the horizontal forces must balance each other.
∑Fy = 0: There is no vertical acceleration, so the vertical forces must balance each other.
Let's start with the ∑Fx equation:
∑Fx = 0
N1 - N2 = 0 (since there are no horizontal forces)
Now, let's move on to the ∑Fy equation:
∑Fy = 0
N1 + N2 - W = 0 (since the normal forces act upward and the weight acts downward)
Substituting the known values into the equation, we get:
N1 + N2 - W = 0
N1 + N2 - 735 = 0
Now, we can use the ratio of sides of the similar right triangles formed by the hands, feet, and body to relate the normal forces:
a/b = N2/W
Substituting the values, we obtain:
40 cm / 95 cm = N2 / 735 N
Simplifying, we find:
N2 = (40 cm / 95 cm) * 735 N = 310.10 N (approximately)
Using the equation N1 - N2 = 0 from the ∑Fx equation, we can solve for N1:
N1 - N2 = 0
N1 = N2
N1 = 310.10 N
Therefore, the normal force exerted by the floor on each hand and on each foot is approximately 310.10 N.