example Δy gravity Δy can

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1 Physic 3 Lecture 5 Main points of today s lecture: Newton s st law: If there is no net force, the velocity of a mass remains constant (neither the magnitude nor the direction of the velocity changes). Newton s nd law: F ma m is the mass of the object. It is proportional p to the number of nucleons (neutrons and protons) in an object. F is the net (total) force acting on the object. It is the vector sum of all forces acting on the object.

2 example A small can is hanging from the ceiling. A rifle is aimed directly at the can, as the figure illustrates. At the instant the gun if fired, the can is released. Ignore air resistance and show that the bullet will always strike the can, regardless of the initial speed of the bullet. Assume that the bullet strikes the can before the can reaches the ground. Δy gravity Δy can θ

3 Reading Quiz. A net force is A. the sum of the magnitudes of all the forces acting on an object. B. the difference between two forces that are acting on an object. C. the vector sum of all the forces acting on an object. D. the force with the largest magnitude acting on an object. Slide 4-7

4 Newton s First Law If there is no net force, the velocity of a mass remains constant (neither the magnitude nor the direction of the velocity changes). Objects at rest feel no net force. Objects in motion with a constant velocity feel no net force. movie

5 Conceptual question Ab ball llis thrown vertically up, its speed slowing under the influence of gravity. Suppose () we film this motion and play the tape backward (so the tape begins with the ball at its highest point and ends with it reaching the point from which it was released), and () we observe the motion of the ball from a frame of reference moving up at the initial speed of the ball. The ball has a downward acceleration g in a) () and (). b) only (). c) only (). d) neither () nor (). () The movie of is the ball run in reverse would look the same as it does run in the forward direction. when it () In the moving frame, the bll ball would look like it is falling downward from rest.

6 Newton s nd law: F Newton s second Law F ma a is the acceleration. mis the mass of the object. It is proportional p to the number of nucleons (neutrons and protons) in an object. F g j is the net (total) force acting on the object. It is the vector sum of all forces acting on the object. Both F and a are vectors with a length and a direction. mis a scaler. It relates the x and y components of the force and acceleration vectors and also their lengths The direction of the force is the same as that of the acceleration. F F x y ma ma x y F ( ma ) + ( ma ) m a + a x Fy ma y ay tan( θ F ) F ma a x y x x x tan( θ ) a y ma Here, θ F and θ a are angles with respect to the x axis.

7 Reading Quiz. Which of the following statements about mass and weight is correct? A. Your mass is a measure of the force gravity exerts on you. B. Your mass is the same everywhere in the universe. C. Your weight is the same everywhere in the universe. D. Your weight is a measure of your resistance of being accelerated. Slide 5-5

8 Inertial mass The unit of mass is kg. The inertial mass m is given by the number of neutrons and protons (nucleons) )in the object. It governs how hard it is to accelerate an object. M N nucleons μ; ; μ.66x0-7 kg

9 Example Abi bicycle has a mass of f3k 3. kg and dits rider has a mass of f87k 8.7 kg. The rider is pumping hard so that a horizontal net force of 9.78 N accelerates them in the westward direction. What is the acceleration? a) 0.03 m/s west b) 0.03 m/s east c) 0.0 m/s west d) 0.0 m/s east e) don t know. m bike m rider F 3. kg 8.7 kg 9.79 N a m bike + F m rider 9.78N 3.kg + 8.7kg 0.03m / s

10 Example A catapult on an aircraft is capable of accelerating a plane from 0 to 56 m/s in a distance of 80 m. Find the magnitude of the average net force that the catapult exerts on a 3300 kg jet. F ma m plane 3300 kg v v aδx 0 v 0 0 v 56 m/s a v v0 v Δ x Δ x Δx 80 m v v F m 0 Δ x ( 56m / s) 80m ( ) 5 3,300.6x0 N

11 quiz A constant tforce is exerted dfor a short ttime interval on a cart tthat t is initially at rest on an air track. This force gives the cart a certain final speed. Suppose we repeat the experiment but, instead of starting from rest, the cart is already moving with constant speed in the direction of the force at the moment we begin to apply the force. After we exert the same constant force for the same short time interval, the increase in the cart s speed a) is equal to two times its initial speed. b) is equal to the square of its initial speed. c) )is equal lto four times its initial iti speed. d) is the same as when it started from rest. e) cannot be determined from the information provided. F v v0 + at v0 + t m v F t m

12 Example A365k 36.5 kg sphere is acted upon by a 33Nf force directed dalong the positive x axis. What is the magnitude and direction of a second force, such that the sphere experiences an acceleration of 3.96 m/s directed at 0 above the positive x axis. F F? m θ? F net [ ] net F F F 36.5kg 3.96m / s 44.5N F 44.5N cos 34.9N ( o ) o net,x 33 N 36.5 kg net,y ( o ) F 44.5Nsin 5.8N F,x Fnet,x F,x 34.9N 33N.9N F F F 5.8N y,y nety net,y,y y 5.8N ( ) ( ) F.9N + 5.8N 5.8N 5.8N θ tan 88.9N o

13 Reading Quiz 3. Which of these is not a force discussed in this chapter? A. The tension force. B. The normal force. C. The orthogonal force. D. The thrust force. Slide 4-

14 Gravitational force: Examples of Forces Objects feel a force from gravity equal to their weights. The weight W of an object equals its mass times the gravitation acceleration g. The weight of an object is less on the moon, but the mass remains the same. The direction of the gravitation force is towards the center of the Earth. 5 kg Wmg T String tension: If a string pulls on an object, we call its force the string tension. It is always in the direction of the string. Strings cannot push. A pulley can redirect the tension force. The net or total force is the vector sum of all forces on an object. The acceleration is inversely proportional to the net force. 5 kg 5 kg T

15 Example A0k kg mass hangs from a string. ti The string ti pulls it so that titi is accelerating upwards with an acceleration of m/s. What is the magnitude of the tension in the string. a) )0N b) 88 N c) 98 N d) 08 N a m/s T mg ma m 0 kg T ma + mg m(a + g) 0 kg T 0kg(0.8m / s ) 08N T

16 Atwood s machine Consider the Atwood machine to the right. The massless string passes over a massless and frictionless pulley. It is under tension T, which we define to be the magnitude of the tension force. By this definition it is a positive number. Choosing up to be positive, what is the net force on mass? a) T-m g b) T+m g c) m g-t d) none of the above m m

17 Choosing up to be positive, what is the net force on mass? a) m g-t b) T+m g c) T-m g d) none of the above Atwood s machine m m

18 From Newton s nd law: T m g T m a m a + m g Atwood s machine Also T m g m a T m a + m g m If m exceeds m, m goes down and m goes up. If m exceeds m, m goes down and m goes up. In either Putting it together: case m a + m g m a + m g Δy Δy ma + mg v v ma + ma mg mg a a ( m ) i m g net gravitational a pulling m m + m up total mass and m down. m

19 Static Equilibrium All objects are at rest and remain so. The net force on any object must vanish. I.e. on an object: F 0 i i Example: Three ropes are arranged so as to support a 4 kg mass as shown below. Determine the tension in each rope. T T 3 A 4 kg ( 0 F ) T i,x 0 T + T cos 60 i T 60 o Solve by considering forces at point A i 0 T + T /; T T / ( 0 ) F 0 T sin 60 T i,y 3 ( 0 ) T T / sin 60.6T 3 3 T3 mg 39N;T 39.N;T 45.3N 453N T.5N