A jumper of mass m=55 kg starts in a crouched position with the jumper's center of mass at a position y1=0 m. Just before the jumper leaves the floor, his or her center of mass is at y2=0.7 m. From another measurement, you have found that the velocity of the jumper's center of mass has only a y component of v2=2.8 m/s as the jumper's feet leave the floor.
assume Ffloor changes while the jumper goes from the crouched to the lift-off position. While the jumper's feet are in contact with the floor assume the force decreases linearly or
Ffloor(y) = -C[y-(y2-y1)],
where y is the changing position of the center of mass, and C is a positive constant. Note that when y=y2-y1 this force goes to zero and, of course, once the jumper's feet leave the floor this force of the floor on the jumper must be zero.
For the initial conditions above, find the maximum value of Ffloor and so the required strength of the floor.
To find the maximum value of Ffloor and the required strength of the floor, we need to determine the force exerted by the floor on the jumper at the lift-off position. We can use the equation given for Ffloor(y) along with the given measurements of the jumper's center of mass positions (y1 and y2) and the velocity component in the y direction (v2).
Let's break down the problem step-by-step:
1. Determine the change in the position of the center of mass (Δy) as the jumper goes from the crouched to the lift-off position:
Δy = y2 - y1
Δy = 0.7 m - 0 m
Δy = 0.7 m
2. Calculate the acceleration of the center of mass (a) using the velocity component (v2) and the change in position (Δy):
v2 = a * t (assuming t = 1 second)
2.8 m/s = a * 1 s
a = 2.8 m/s^2
3. Determine the net force acting on the jumper at the lift-off position using Newton's second law:
Fnet = m * a
Fnet = 55 kg * 2.8 m/s^2
Fnet = 154 N
4. Substitute the value of Fnet into the equation for Ffloor(y) to find the maximum value of Ffloor:
Ffloor(y) = -C * [y - (y2 - y1)]
Ffloor(y) = -C * [y - (0.7 m - 0 m)]
Ffloor(y) = -C * (y - 0.7 m)
At the lift-off position, y = y2, so we plug in that value:
Ffloor(y2) = -C * (y2 - 0.7 m)
Ffloor(y2) = -C * (0.7 m - 0.7 m)
Ffloor(y2) = -C * 0 m
Ffloor(y2) = 0 N
From this calculation, we find that the force exerted by the floor on the jumper at the lift-off position (Ffloor(y2)) is zero. This means that the required strength of the floor is zero.
However, it's important to note that the initial assumption in the problem states that the force decreases linearly. This suggests that the maximum value of Ffloor would occur at the crouched position (y = y1). To find the maximum value of Ffloor, we need more information or clarification in the problem statement.
To find the maximum value of Ffloor, we need to analyze the forces acting on the jumper at various positions.
Let's consider the forces acting on the jumper when he/she is at y = y2 (the lift-off position). At this point, the only vertical force acting on the jumper is the gravitational force.
Fgravity = mg
where m is the mass of the jumper and g is the acceleration due to gravity.
Next, let's calculate the net force acting on the jumper when he/she is at y = y2. Since the jumper is leaving the floor, the net vertical force must be zero.
Fnet = Fgravity + Ffloor = 0
Substituting the expression for Ffloor in terms of y:
mg - C(y - (y2 - y1)) = 0
Now, let's solve for y to find the maximum value of Ffloor.
mg - Cy + C(y2 - y1) = 0
Simplifying the equation:
Cy = mg + C(y2 - y1)
Cy = mg + Cy2 - Cy1
Rearranging the equation:
Cy - Cy2 = mg - Cy1
Factoring out C:
C(y - y2) = mg - Cy1
Dividing both sides by (y - y2):
C = (mg - Cy1) / (y - y2)
To find the maximum value of Ffloor, we need to minimize the denominator (y - y2). Therefore, the maximum value of Ffloor occurs when (y - y2) approaches zero.
Substituting (y - y2) = 0 into the equation:
Cmax = (mg - Cy1) / 0
However, it is not possible to divide by zero. Therefore, the maximum value of Ffloor is undefined, indicating that there is no required strength of the floor to support the jumper.
This implies that there is a gap between leaving the floor and reaching the lift-off position where the jumper is momentarily airborne. During this period, the jumper relies on the initial upward force from the floor and the acceleration due to gravity to lift off.