Static Equilibrium Gravitation

Transcription

1 Static Equilibrium Gravitation Lana Sheridan De Anza College Dec 6, 2017

2 Overview One more static equilibrium example Newton s Law of Universal Gravitation gravitational potential energy little g

3 Example Slipping Ladder ticity A uniform ladder of length l, rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is µ s = Find the minimum angle θ min at which the ladder does not slip., vertical wall (Fig. nt of static friction ind the minimum S P S n mbed. Do you want er and the surface dder stay up? Simul surface. Does the model it as a rigid a u O b S f s u S mg Figure 12.9 (Example 12.3) (a) A uniform ladder at rest, leaning against a smooth wall. The ground is rough. (b) The forces on the ladder.

4 Example Slipping Ladder Important point: the wall is described as smooth and we are not told anything about a coefficient of friction with the wall. neglect friction between the ladder and the wall. The only force the wall exerts on the ladder is a normal force.

5 Example Slipping Ladder Important point: the wall is described as smooth and we are not told anything about a coefficient of friction with the wall. neglect friction between the ladder and the wall. The only force the wall exerts on the ladder is a normal force. Forces in x-direction: Forces in y-direction: f s P = 0 n mg = 0 Torques. Pick the point O for simplicity: Pl sin θ mg l 2 cos θ = 0

6 Example Slipping Ladder From torque equation: tan θ = mg 2P When is θ at minimum? When tan θ is at minimum, P at maximum. P = f s P max = f s,max = µ s mg tan θ = mg 2P 1 = 2µ s = 51

7 Gravitation The force that massive objects exert on one another. Newton s Law of Universal Gravitation F G = Gm 1m 2 r 2 for two objects, masses m 1 and m 2 at a distance r. G = Nm 2 kg 2.

8 r to generate a value for G., each of mass m, fixed to the er Gravitation or thin metal wire as illusf mass M, are placed near the larger spheres causes the rod brium orientation. The angle eam reflected from a mirror.1 is often referred to as an e varies as the inverse square examples of this type of force in vector form by defining a rected from particle 1 toward is (13.3) 5 k/x, where k is a constant. A direct pro- Figure 13.1 Cavendish apparatus for measuring gravitational forces. S F 21 m 1 Consistent with Newton s S S third law, F 21 F 12. rˆ12 r S F 12 m 2 Figure 13.2 The gravitational force between two particles is attractive. The unit vector r^12 is directed from particle 1 toward particle 2. r 2 ˆr 1 2 F G,1 2 = Gm 1m 2 for two objects, masses m 1 and m 2 at a distance r. G = Nm 2 kg 2.

9 vitational forces on the cue ball using Equation Once these R E Example 13.1: The Gravitational Force between Small Masses is Small rth s mass and R E its radius. This force is directed toward the planet has two moons of equal mass. Moon 1 is in a circular oon 2 is in a circular orbit of radius 2r. What is the magnitude al force exerted by the planet on Moon 2? (a) four times as large (b) twice as large as that on Moon 1 (c) equal to that on Moon 1 s that on Moon 1 (e) one-fourth as large as that on Moon 1 Three kg billiard balls are placed on a table at the corners of a right triangle. The sides of the triangle are of lengths a = m, b = m, and c = m. Calculate the gravitational force vector on the cue ball (designated m 1 ) resulting from the other two balls as well as the magnitude and direction of this force. y m 2 a c t the corners of a right triangle re of lengths a m, b 5 al force vector on the cue ball well as the magnitude and direcall is e. We ward es as ion of Figure 13.3 (Example 13.1) The resultant gravitational force acting on the cue ball is the vector sum S F21 1 S F31. S S F F21 S F u 31 x m 1 b m 3

10 Example 13.1 Remember, G = N m 2 kg 2 : F G,1 2 = Gm 1m 2 r 2 ˆr 1 2

11 Example 13.1 Remember, G = N m 2 kg 2 : F G,1 2 = Gm 1m 2 r 2 ˆr 1 2 F 2 1 = Gm 2m 1 r 2 = Gm 2m 1 j a 2 ˆr 2 1 = j N F 3 1 = Gm 3m 1 r 2 = Gm 3m 1 i b 2 ˆr 3 1 = i N

12 Example 13.1 Remember, G = N m 2 kg 2 : F G,1 2 = Gm 1m 2 r 2 ˆr 1 2 F 2 1 = Gm 2m 1 r 2 So, = Gm 2m 1 j a 2 ˆr 2 1 = j N F 3 1 = Gm 3m 1 r 2 = Gm 3m 1 i b 2 ˆr 3 1 = i N F net = F F 3 1 = ( i j) N Very small! F net = N with θ = 29.3

13 Gravitational Potential Energy Remember from Chapter 7, F x = du dx ; U = F (x) dx Since W = F (r) dr, this tells us that the work done by gravity on an object is equal to minus the change in potential energy.

14 Gravitational Potential Energy Remember from Chapter 7, F x = du dx ; U = F (x) dx Since W = F (r) dr, this tells us that the work done by gravity on an object is equal to minus the change in potential energy. rf U = r i rf F(r) dr = Gm 1m 2 r i r 2 dr ( 1 = Gm 1 m 2 1 ) r f r i

15 Gravitational Potential Energy It is useful to pick a reference point to set the scale for gravitational potential energy. What would be a good point?

16 Gravitational Potential Energy It is useful to pick a reference point to set the scale for gravitational potential energy. What would be a good point? Infinite distance! For r i =, U(r i ) = 0. Then we can define: U(r) = Gm 1m 2 r This will always be a negative number.

17 Gravitational Potential Energy (13.14) center of the Earth particles inside the U is always negative th system, a similar is, the gravitational es m 1 and m 2 sepa- (13.15) for any pair of par- 1/r 2. Furthermore, and we have chosen finite. Because the do positive work to y the external agent Gravitational potential energy of the Earth particle U(r) = Gm system 1m 2 r O GM E m R E U M E Earth The potential energy goes to zero as r approaches infinity. R E Figure Graph of the grav- r

18 Acceleration due to Gravity This force in that it gives objects weight, F g. For an object of mass m near the surface of the Earth: and where F g = mg g = GM E R 2 E M E = kg is the mass of the Earth and R E = m is the radius of the Earth. The force F g acts downwards towards the center of the Earth.

19 Acceleration due to Gravity The acceleration due to gravity, g, can vary with height! F G = GM ( ) Em GME r 2 = m r 2 = mg Depends on r the distance from the center of the Earth. Suppose an object is at height h above the surface of the Earth, then: g decreases as h increases. g = GM E (R E + h) 2

20 Acceleration due to Gravity The acceleration due to gravity, g, can vary with height! F G = GM ( ) Em GME r 2 = m r 2 = mg Depends on r the distance from the center of the Earth. Suppose an object is at height h above the surface of the Earth, then: g decreases as h increases. g = GM E (R E + h) 2 g is the not just the acceleration due to gravity, but also the magnitude of the gravitational field.

21 Summary static equilibrium practice Newton s Law of Universal Gravitation gravitational potential energy little g 4th Collected Homework! due tomorrow. (Uncollected) Homework Serway & Jewett, PREV: Ch 12, onward from page 400. Questions: Section Qs 3, 11, 15, 19, 23, 25 Ch 13, onward from page 410. Questions: Section Qs 3, 9, 11, 15, 31, 33, 35