1. Universal Law of Gravitation, Normal Force, Apparent Weight SPH3U

2. Newton’s work The acceleration of gravity must be caused by a net force, according to Newton’s 2 nd Law. Every particle of matter in the universe exerts an attractive force on every other particle. The constant “G” is called the universal gravitational constant which is 6.67259 x 10 -11 Nm 2 /kg 2 G” was first measured experimentally by Henry Cavendish (1731-1810) more than a century after Newton proposed his theory! The distances between the objects must be between their centres, and small masses have negligible radii. “

3. Example Find the magnitude of the gravitational force exerted between a biologist (m = 55.0 kg) and a physicist (m = 85.0 kg) when they are 2.00 m apart. If a direction was required, the best choices would be “towards centre” (or tc) or “attractive”, as we don’t know the force’s direction. (Later, an FBD for some questions, yields a direction.)

4. Example 2 Find the weight of the 11 600 kg Hubble Space Telescope on the Earth’s surface AND in its orbit, 596 km above the Earth’s surface. The radius of the Earth is 6.37 x 10 6 m.

5. Example 3 The acceleration due to gravity on the moon is 1/6 th that of Earth. What is the weight of a 1.00 kg object on the moon? On Earth g = 9.81 ms -2 F g = mg is safe to use for objects on the surface. On the moon F g = 1.00 kg (9.81/6)ms -2 F g = 1.64 N [down]

6. Example 4 What’s the mass and weight of a 12.0 kg mass in space? The mass is an intrinsic property of matter. It stays at 12.0 kg. The weight is 0 N if all mass are infinitely far away!

7. Example 5 You stand on a frozen lake watching a speeding ice boat approach you. Being a keen physicist, you wonder what force would be needed to decelerate the boat. Then you wonder if you were on another planet, (where the weight of the speedboat was doubled), what force would be needed to decelerate the boat on the planet? Would the force needed on the planet be the same, less, or greater than the force on Earth? (Ignore all frictions). The force required to decelerate it depends on mass and acceleration, according to Newton’s 2 nd Law: F net = ma. Mass has not changed, so the force required to decelerate it is the same! The boat has a greater weight due to the planet’s stronger attraction, which is misleading.

8. Normal Force In many cases, an object is in contact with a surface. The contact implies a force acts on the object. The perpendicular part of this force is called the Normal force, and according to Newton’s 3 rd Law, is a reaction force to gravity. A good analogy is when you sit on a bed, compressing the springs. These springs exert an upward force to support your weight, when you come to rest (not moving). In a similar way, atomic springs push up on the block resting on the table. (The normal force is a result of the electric force between particles in the atoms of the block and table.)

9. Normal Force The magnitude of the normal force indicates how hard the two objects press on each other. If an object is resting on a horizontal table, and there are no other vertical forces acting, then |F g |= |F N |. If you were to push down on the block, the normal force would increase. |F N | = |F g + F a | Ex) F g = 15 N [down], F a = 11 N [down] so F N = 26 N [up] If you pulled up on the block (without lifting it off), the normal force would reduce. |F N | = |F g - F a |, until the applied force equals the weight.

10. Circus Question In a circus act, a man balances a woman doing a handstand on his head. The woman weighs 490.0 N [down] and the man’s head weighs 50.0 N [down]. The 7 th cervical vertebra in the spine supports all weight above the shoulders. Find the normal force this vertebra exerts on the man’s head and neck a) before the act b) during the act. FBD’s help to visualize:

11. Circus Question a) Before the act, the man’s head has a downward weight of 50.0 N and if the man’s head is to remain at rest, the upward force must also be 50.0 N.  F N = 50.0 N [up] b) The total downward force must be balanced by an upward force, if the man’s head remains at rest.  F N = 50.0 N + 490.0 N = 540.0 N [up]

12. Apparent Weight The weight we experience is a result of the normal force pushing up on our feet. Our apparent weight (what we experience) may alter if the normal force changes. Elevators are good examples of this situation. When an elevator accelerates upward, there must be an applied upward force greater than the weight of the elevator. The applied upward force adds to the normal force on our feet, making us feel heavier. We feel lighter in an elevator that accelerates downwards, in a similar way.

13. Apparent Weight The normal force indicates our apparent weight as the vector sum of our true weight F g and the net force F net. F N = mg + ma F N is the magnitude of the normal force exerted on the person, which must also be the downward force of the person: the apparent weight (using Newton’s 3 rd Law). If the acceleration is 0 (at rest or uniform motion) then the apparent weight is the true weight F g. If the elevator accelerates upwards, then F N = m (g + a) If the elevator accelerates downwards then F N = m (g – a)

14. Examples A 75.0 kg biologist is so bored with his science, he rides elevators all day. What is the apparent weight of the biologist if: a) The elevator moves upwards at 2.00 m/s 2 ? F N = m(g + a) = 75.0 kg (9.81 m/s 2 + 2.00 m/s 2 ) = 886 N [up] b) The elevator moves upwards at 3.00 m/s? F N = m (g + a) = 75.0 kg (9.81 m/s 2 + 0) = 736 N [up] c) The elevator moves downwards at 2.00 m/s 2 ? F N = m (g – a) = 75.0 kg (9.81 m/s 2 – 2.00 m/s 2 ) = 586 N [up]