A man doing push-ups pauses in the position shown in the figure . His mass = 68 kg.
Image: img145.imageshack.us/img145/7130/picture1n.png
a= 40 cm
b= 95 cm
c= 30 cm
Determine the normal force exerted by the floor on each hand.
Determine the normal force exerted by the floor on each foot.
To determine the normal force exerted by the floor on each hand, we need to analyze the forces acting on the man's body.
1. First, let's identify the forces acting on the man's upper body. We have the weight acting downwards (mg), the normal force exerted by the floor (N1), and the force exerted by the man's arms (F1) during the push-up.
2. To start, let's calculate the weight of the man. Given that his mass is 68 kg and the acceleration due to gravity is approximately 9.8 m/s^2, the weight can be calculated as follows:
weight = mass * acceleration due to gravity
weight = 68 kg * 9.8 m/s^2
weight = 666.4 N
3. Since the man is in a static position, the sum of the forces in the vertical direction must be zero. Therefore, the normal force exerted by the floor must be equal and opposite to the weight of the man:
N1 = weight = 666.4 N
4. Since the man is in a static position, the sum of the forces in the horizontal direction must also be zero. Therefore, the force exerted by the man's arms must be equal and opposite to the horizontal component of the weight:
F1 = mg * sin(a)
F1 = 68 kg * 9.8 m/s^2 * sin(40°)
F1 ≈ 427.2 N
Therefore, the normal force exerted by the floor on each hand is approximately 427.2 N.
To determine the normal force exerted by the floor on each foot, we need to analyze the forces acting on the man's lower body.
1. We have the weight acting downwards (mg), the normal force exerted by the floor (N2), and the force exerted by the man's legs (F2) during the push-up.
2. Since the sum of the forces in the vertical direction must be zero, the normal force exerted by the floor must be equal and opposite to the weight of the man:
N2 = weight = 666.4 N
3. Since the sum of the forces in the horizontal direction must be zero, the force exerted by the man's legs must be equal and opposite to the horizontal component of the weight:
F2 = mg * sin(c)
F2 = 68 kg * 9.8 m/s^2 * sin(30°)
F2 ≈ 332.3 N
Therefore, the normal force exerted by the floor on each foot is approximately 332.3 N.
To determine the normal force exerted by the floor on each hand and foot, we can consider the equilibrium of forces acting on the man's body.
First, let's label the forces acting on each hand and foot:
For the hand:
- The weight of the man acting downward (mg)
- The normal force exerted by the floor on the hand acting upward (N_hand)
For the foot:
- The weight of the man acting downward (mg)
- The normal force exerted by the floor on the foot acting upward (N_foot)
Now, let's calculate the normal force exerted by the floor on each hand and foot.
For the hand:
In order for the man's hand to be in equilibrium, the sum of the forces acting vertically (in the y-direction) must be zero.
Therefore, we can write the equation:
N_hand - mg = 0
Solving for N_hand, we find:
N_hand = mg
For the foot:
Similarly, in order for the man's foot to be in equilibrium, the sum of the forces acting vertically (in the y-direction) must be zero.
Therefore, we can write the equation:
N_foot - mg = 0
Solving for N_foot, we find:
N_foot = mg
Now, let's substitute the mass of the man (68 kg) into the equations to find the values of N_hand and N_foot.
For the hand:
N_hand = mg = 68 kg * 9.8 m/s^2 = 666.4 N (rounded to one decimal place)
For the foot:
N_foot = mg = 68 kg * 9.8 m/s^2 = 666.4 N (rounded to one decimal place)
Therefore, the normal force exerted by the floor on each hand is approximately 666.4 N, and the normal force exerted by the floor on each foot is also approximately 666.4 N.
Let Fl be the force upward on the hands, Fr be the force on his toes.
Fr+Fl=weight
Now sum moments about any point, I will sum them about the feet. Clockwise moments are postive. -weight*b+Fl*(a+b)=0
you have two equations, two unknowns (Fl,Fr). solve. Now for each hand,divide Fl by 2, and for each foot, divide Fr by 2.