Question

What is the magnitude and direction of the force required from your legs to jump a certain horizontal distance, R? Express the result in terms of your weight, W = mg. Assume that you project yourself into the jump by lowering your body and forcefully straightening your legs, so the force that launches you is applied over the extent of the crouching position, d. Also assume that you angle the direction of the applied force from your legs in such a way that the net force points in the optimal direction, the direction that maximizes the range, R, of your jump, which we know to be 45 degrees. The following prompts will help lead you to an expression for the applied force.

-Estimate reasonable numeric values for R and d.

-Draw a free-body-diagram to depict the forces on the jumper at takeoff.

-Express the acceleration required to achieve the takeoff velocity in terms of R and d.

-Analyze the forces and use the acceleration above to solve for the applied force in terms of the weight, W = mg.

-Can you solve for the applied force and its direction before substituting any numeric quantities? What is gained by doing so?

- What can you do to make yourself go further? Why would this be helpful

Answer #1

When you jump, you start by crouching down a bit, which
lowers your center of mass to a distance d ≈ 40 cm below
where it normally is. You then exert a downward contact force on
the ground equal to about 2.5 times your weight (mg); we
may assume that this force is constant while it acts. After your
center of mass rises past its usual location, you leave the ground
with an initial velocity v, and you reach a...

Consider a 6-legged water-treader with a body mass of 18 mg
(milligrams) and perfectly-non-wetting legs of total combined
length 20 mm standing on a freshwater pond. The body length is 9 mm
and the body density is ?ρ = 1200 kg/m33. Consider the mass of the
legs to be negligible.
The equation for the upwards surface-tension force that prevents
the water-treader from sinking is given by
???=2??cos?,Fst=2γLcosθ,
where ?θ is the angle of contact between the water and the legs,...

Consider a 6-legged water-treader with a body mass of 18 mg
(milligrams) and perfectly-non-wetting legs of total combined
length 20 mm standing on a freshwater pond. The body length is 10
mm and the body density is ? = 1200 kg/m3. Consider the
mass of the legs to be negligible.
The equation for the upwards surface-tension force that prevents
the water-treader from sinking is given by
F = 2γLcosθ
where θ is the angle of contact between the water and...

Design a "bungee jump" apparatus for adults. A bungee jumper falls
from a high platform with two elastic cords tied to the ankles. The
jumper falls freely for a while, with the cords slack. Then the
jumper falls an additional distance with the cords increasingly
tense. Assume that you have cords that are 12
m long, and that the cords stretch in the jump an additional
19
m for a jumper whose mass is 120
kg, the heaviest adult you...

1. For a stationary ball of mass m = 0.200 kg hanging from a
massless string, draw arrows (click on the “Shapes” tab) showing
the forces acting on the ball (lengths can be arbitrary, but get
the relative lengths of each force roughly correct). For this case
of zero acceleration, use Newton’s 2nd law to find the
magnitude of the tension force in the string, in units of Newtons.
Since we will be considering motion in the horizontal xy plane,...

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INNOVATION
Deep Change: How Operational Innovation Can Transform Your
Company
by
Michael Hammer
From the April 2004 Issue
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8.95
In 1991, Progressive Insurance, an automobile insurer based in
Mayfield Village, Ohio, had approximately $1.3 billion in sales. By
2002, that figure had grown to $9.5 billion. What fashionable
strategies did Progressive employ to achieve sevenfold growth in
just over a decade? Was it positioned in a high-growth industry?
Hardly. Auto insurance is a mature, 100-year-old industry...

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