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Worksheet 308

Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational.

(a) Calculate the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child).

(b) Calculate the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29e+11 m away. How does the force of Jupiter on the baby compare to the force of the father on the baby?

Father’s gravitational force on the baby is:

Newton's law of universal gravitation

Jupiter’s gravitational force on the baby is:

Newton's law of universal gravitation
Division

(c) What should be the father’s weight, so that he exerts the same force on the baby as that of Jupiter? **
**this section is not included in the Reference material

Newton's law of universal gravitation

Discussion

Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics

Dedicated to little Konstantinos

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Worksheet 306

Calculate the force the biceps muscle must exert to hold the forearm and its load as shown in the figure below, and compare this force with the weight of the forearm plus its load. You may take the data in the figure to be accurate to three significant figures.

(a) The figure shows the forearm of a person holding a book. The biceps exert a force FB to support the weight of the forearm and the book. The triceps are assumed to be relaxed. (b) Here, you can view an approximately equivalent mechanical system with the pivot at the elbow joint

Strategy

There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is FB, that of the elbow joint is FE, that of the weights of the forearm is wa , and its load is wb. Two of these are unknown FB, so that the first condition for equilibrium cannot by itself yield FB . But if we use the second condition and choose the pivot to be at the elbow, then the torque due to FE is zero, and the only unknown becomes FB .

Solution

The torques created by the weights are clockwise relative to the pivot, while the torque created by the biceps is counterclockwise; thus, the second condition for equilibrium (net τ = 0) becomes

Force (Newton's second law)
Torque
Force (Newton's second law)
Torque

Note that sin θ = 1 for all forces, since θ = 90º for all forces. This equation can easily be solved for FB in terms of known quantities,yielding. Entering the known values gives

Mechanical equilibrium - 3=3 Torque example

which yields

Torque

Now, the combined weight of the arm and its load is known, so that the ratio of the force exerted by the biceps to the total weight is

Division

Discussion

This means that the biceps muscle is exerting a force 7.38 times the weight supported.

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics