The drawing shows a person (weight W = 580 N, L1 = 0.849 m, L2 = 0.410 m) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position.
To find the normal force exerted by the floor on each hand and each foot, we can use the principles of static equilibrium.
Let's denote the normal force exerted by the floor on each hand as Fh1 and Fh2, and the normal force exerted by the floor on each foot as Ff1 and Ff2.
From the given information, we have:
Weight of the person (W) = 580 N
Length from the hands to the person's center of gravity (L1) = 0.849 m
Length from the feet to the person's center of gravity (L2) = 0.410 m
In static equilibrium, the sum of the torques about any point must be zero. We can choose the center of gravity as the point about which we calculate the torques.
The torque due to weight about the center of gravity is given by:
Torque = Weight x Length
For each hand, the torque due to weight can be calculated as:
Torqueh1 = Fh1 x L1
Torqueh2 = Fh2 x L1
Similarly, for each foot, the torque due to weight can be calculated as:
Torquef1 = Fh1 x L2
Torquef2 = Fh2 x L2
Since the person is in static equilibrium, the sum of the torques about the center of gravity must be zero:
Torqueh1 + Torqueh2 + Torquef1 + Torquef2 = 0
Substituting the expressions for torque and weight:
(Fh1 x L1) + (Fh2 x L1) + (Ff1 x L2) + (Ff2 x L2) = 0
Simplifying the equation:
Fh1 + Fh2 + Ff1 + Ff2 = (Weight x L1) / L2
Now, we can use the fact that the sum of the vertical forces must balance the weight of the person:
Fh1 + Fh2 + Ff1 + Ff2 = Weight
Substituting the given weight:
Fh1 + Fh2 + Ff1 + Ff2 = 580 N
Now, we have two equations with four unknowns. However, we can use the fact that the person is doing push-ups, which means that the normal force on each hand is equal and the normal force on each foot is equal.
So, we can write:
Fh1 = Fh2 = Fh (let's call this common value as Fh)
Ff1 = Ff2 = Ff (let's call this common value as Ff)
Substituting these values into the equation:
Fh + Fh + Ff + Ff = 580 N
2Fh + 2Ff = 580 N
Dividing both sides of the equation by 2:
Fh + Ff = 290 N
Now, we have two equations with two unknowns:
Fh + Ff = 290 N
Fh x L1 + Fh x L1 + Ff x L2 + Ff x L2 = 0
From the first equation, we can express Fh in terms of Ff:
Fh = 290 N - Ff
Substituting this expression into the second equation:
(290 N - Ff) x L1 + (290 N - Ff) x L1 + Ff x L2 + Ff x L2 = 0
Expanding and simplifying the equation:
(580 N - 2Ff) x L1 + 2Ff x L2 = 0
614.68 N - 2Ff = 0
2Ff = 614.68 N
Ff = 307.34 N
Finally, substituting the value of Ff back into the first equation:
Fh + Ff = 290 N
Fh + 307.34 N = 290 N
Fh = -17.34 N (negative sign indicates downward force)
Therefore, the normal forces exerted by the floor on each hand are approximately -17.34 N (downward force) and the normal forces exerted by the floor on each foot are approximately 307.34 N (upward force).
To find the normal force exerted by the floor on each hand and each foot, we can analyze the forces acting on the person in this position.
Let's start by looking at the forces acting on the entire person. We have the weight of the person, W, acting downwards (580 N). In this position, the person is in equilibrium, so the sum of the vertical forces must be zero.
Let's consider the forces acting on each hand first. There are two forces acting on each hand: the weight of the person and the normal force exerted by the floor. Both of these forces act vertically. Since the person is in equilibrium, the sum of the vertical forces acting on each hand must be zero.
Using this information, we can set up an equation for each hand:
Weight of the person = Normal force exerted by the floor on each hand
Now let's consider the forces acting on each foot. Similar to the hands, there are two forces acting on each foot: the weight of the person and the normal force exerted by the floor. Again, the sum of the vertical forces acting on each foot must be zero.
Using this information, we can set up an equation for each foot:
Weight of the person = Normal force exerted by the floor on each foot
To find the normal forces exerted by the floor on each hand and each foot, we can use the equations we derived above. Let's calculate them:
For the hands:
Normal force on each hand = Weight of the person / 2
For the feet:
Normal force on each foot = Weight of the person / 2
Substituting the given weight of the person (W = 580 N) into the equations:
Normal force on each hand = 580 N / 2 = 290 N
Normal force on each foot = 580 N / 2 = 290 N
Therefore, the normal force exerted by the floor on each hand and each foot, assuming the person holds this position, is 290 N.