The drawing shows a person (weight W = 581 N, L1 = 0.843 m, L2 = 0.404 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.
force on each hand
N
force on each foot
N
without the drawing, ....
We have vertical and rotational equilibrium.
All vertical forces are balanced:
2•N1 +2•N2 –W = 0.
(N1+N2) = W/2.
The sum of torques around the feet are
W•L1 – 2•N2(L1+L2) = 0.
581•0.843 =2•N2• (0.843+0.404),
N2 =581•0.843/2• (0.843+0.404) =
=196.4 N.
N1+N2 =581/2= 290.5
N2 = 290.5-196.4=94.1 N
To find the normal force exerted by the floor on each hand and each foot, you can start by considering the forces acting on the person in the push-up position.
1. Force on each hand:
The person's weight (W = 581 N) is acting downward. Since the person is in equilibrium, the sum of the vertical forces must be zero. So, the normal force exerted by the floor on each hand will balance the weight of the person.
Therefore, the force on each hand is equal to half of the person's weight, as each hand supports an equal portion of the weight.
Force on each hand = W/2 = 581 N / 2 = 290.5 N
2. Force on each foot:
Similarly, the normal force exerted by the floor on each foot will also balance the weight of the person.
However, since the person's feet are closer together compared to their hands, the force on each foot will be less than half of the person's weight.
To find the force on each foot, you can use the principle of moments. The sum of the moments about any point (for example, the point between the feet) must be zero in the equilibrium position.
Moment due to weight = Moment due to force on left foot + Moment due to force on right foot
(W * L1) = (Fleft * L2) + (Fright * L2)
By substituting the given values, the equation becomes:
(581 N * 0.843 m) = (Fleft * 0.404 m) + (Fright * 0.404 m)
From the equation, we can find the force on each foot:
Force on each foot = (W * L1) / (2 * L2)
= (581 N * 0.843 m) / (2 * 0.404 m)
≈ 706.29 N
Therefore, the force on each foot is approximately 706.29 N.
So, the answers are:
Force on each hand ≈ 290.5 N
Force on each foot ≈ 706.29 N
To find the normal force exerted by the floor on each hand and foot, we need to consider the forces at play in this situation.
In a push-up position, there are three forces acting on the person: the weight (W) of the person acting downwards, and the normal forces on each hand and foot acting upwards.
To find the normal force on each hand, we can consider the vertical forces acting on the person's upper body. Since the person is in equilibrium (not accelerating), the sum of the vertical forces must be zero.
Let's calculate the total vertical force acting on the person's upper body:
Total vertical force = Force on each hand + Force on each foot - Weight of the person
Since the person's hands and feet are in contact with the floor, the normal forces on each hand and foot are equal in magnitude and opposite in direction. Therefore, we have:
Force on each hand = Force on each foot = N (let's assume this value for now)
Now, let's substitute this value into the equation to find the force on each hand:
Total vertical force = 2 * N - W
Since the person is holding the push-up position, the total vertical force is equal to zero:
0 = 2 * N - W
Now, we can solve this equation for N (force on each hand):
2 * N = W
N = W / 2
Substituting the given values into the equation:
N = 581 N / 2 = 290.5 N
Therefore, the normal force exerted by the floor on each hand is 290.5 N.
Similarly, the normal force exerted by the floor on each foot is also 290.5 N.