Force and Newton’s Laws Free body diagram with forces drawn from center of mass (fine if objects behave as point masses) Free body diagram with forces drawn from point of applica>on (necessary if rota>on is a possibility) Force and Newton’s Laws FREE BODY DIAGRAMS 2D Spheres: Draw a free body diagram for each “sphere.” Bosun Chair: Draw a free body diagram for the person, the chair, and the person-­‐chair system. Force and Newton’s Laws 5.3 Newton’s First Law
Newton’s First Law: If no force acts on a body, the body’s velocity cannot change; that is, the body cannot accelerate. If the body is at rest, it stays at rest. If it is moving, it con>nues to move with the same velocity (same magnitude and same direc+on). 5.4 Force
Ø A force is measured by the accelera>on it produces. Ø Forces have both magnitudes and direc>ons. Ø When two or more forces act on a body, we can find their net, or resultant force, by adding the individual forces vectorially. Ø The net force ac>ng on a body is represented with the vector symbol !
F
Ø Newton’s First Law: net
If no net force acts on a body , the body’s velocity cannot change; that is, the body cannot accelerate 5.4 Force
The force that is exerted on a standard mass of 1 kg to produce an accelera>on of 1 m/s2 has a magnitude of 1 newton (abbreviated N) Iner>al Reference Frames An iner>al reference frame is one in which Newton’s laws hold. (a) The path of a puck sliding from the north pole as seen from a sta+onary point in space. Earth rotates to the east. (b) The path of the puck as seen from the ground. If a puck is sent sliding along a short strip of fric>onless ice—the puck’s mo>on obeys Newton’s laws as observed from the Earth’s surface. If the puck is sent sliding along a long ice strip extending from the north pole, and if it is viewed from a point on the Earth’s surface, the puck’s path is not a simple straight line. The apparent deflec>on is not caused by a force, but by the fact that we see the puck from a rota>ng frame. In this situa>on, the ground is a noniner>al frame. 5.6: Newton’s second law:
The net force on a body is equal to the product of the body’s mass and its accelera>on. In component form, The accelera>on component along a given axis is caused only by the sum of the force components along that same axis, and not by force components along any other axis. 5.7: Some particular forces
Gravita>onal Force: A gravita>onal force on a body is a certain type of pull that is directed toward a second body. Suppose a body of mass m is in free fall with the free-­‐fall accelera>on of magnitude g. The force that the body feels as a result is: Fg = m(g) or Fg = mg. The weight, W, of a body is equal to the magnitude Fg of the gravita+onal force on the body. W = mg (weight), 5.7: Some particular forces
Normal Force: When a body presses against a surface, the surface (even a seemingly rigid one) deforms and pushes on the body with a normal force, FN, that is perpendicular to the surface. In the figure, forces Fg and FN and are the only two forces on the block and they are both ver>cal. Thus, for the block we can write Newton’s second law for a posi>ve-­‐upward y axis, (Fnet, y= may), as: Fig. 5-­‐7 (a) A block res>ng on a table experiences a normal force perpendicular to the tabletop. (b) The free-­‐body diagram for the block. for any ver>cal accelera>on ay of the table and block 5.7: Some particular forces
Fric>on If we either slide or a_empt to slide a body over a surface, the mo>on is resisted by a bonding between the body and the surface. The resistance is considered to be single force called the fric>onal force, f. This force is directed along the surface, opposite the direc>on of the intended mo>on. 5.7: Some particular forces
Tension When a cord is a_ached to a body and pulled taut, the cord pulls on the body with a force T directed away from the body and along the cord. Fig. 5-­‐9 (a) The cord, pulled taut, is under tension. If its mass is negligible, the cord pulls on the body and the hand with force T, even if the cord runs around a massless, fric>onless pulley as in (b) and (c). 5.8: Newton’s Third Law
When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direc>on. •  The minus sign means that these two forces are in opposite direc>ons •  The forces between two interac>ng bodies are called a third-­‐law force pair. 5.9: Applying Newton’s Laws
Sample problem Figure 5-­‐12 shows a block S (the sliding block) with mass M =3.3 kg. The block is free to move along a horizontal fric>onless surface and connected, by a cord that wraps over a fric>onless pulley, to a second block H (the hanging block), with mass m 2.1 kg. The cord and pulley have negligible masses compared to the blocks (they are “massless”). The hanging block H falls as the sliding block S accelerates to the right. Find (a) the accelera>on of block S, (b) the accelera+on of block H, and (c) the tension in the cord. Key Ideas: 1.  Forces, masses, and accelera>ons are involved, and they should suggest Newton’s second law of mo>on: 2.  The expression !
! is a vector equa>on, Fso =wme acan it as three ! write !
component equa>ons. F = ma
3.  Iden>fy the forces ac>ng on each of the bodies and draw free body diagrams. 5.9: Applying Newton’s Laws
Sample problem In Fig. a, a cord pulls on a box of sea biscuits up along a fric>onless plane inclined at θ= 30°.The box has mass m =5.00 kg, and the force from the cord has magnitude T =25.0 N. What is the box’s accelera>on component a along the inclined plane? For convenience, we draw a coordinate system and a free-­‐body diagram as shown in Fig. b. The posi>ve direc>on of the x axis is up the plane. Force from the cord is up the plane and has magnitude T=25.0 N. The gravita>onal force is downward and has magnitude mg =(5.00 kg)(9.8 m/s2) =49.0 N. Also, the component along the plane is down the plane and has magnitude mg sin θ as indicated in the following figure. To indicate the direc>on, we can write the down-­‐the-­‐plane component as -­‐mg sin θ. Using Newton’s Second Law, we have : which gives: The posi>ve result indicates that the box accelerates up the plane. 6.3 Properties of friction
Property 1. If the body does not move, then the sta>c fric>onal force and the component of F that is parallel to the surface balance each other. They are equal in magnitude, and is fs directed opposite that component of F. Property 2. The magnitude of has a maximum value fs,max that is given by where µs is the coefficient of sta5c fric5on and FN is the magnitude of the normal force on the body from the surface. If the magnitude of the component of F that is parallel to the surface exceeds fs,max, then the body begins to slide along the surface. Property 3. If the body begins to slide along the surface, the magnitude of the fric>onal force rapidly decreases to a value fk given by where µk is the coefficient of kine5c fric5on. Thereaier, during the sliding, a kine5c fric5onal force fk opposes the mo>on. 6.2 Frictional Force: motion of a crate with applied forces
There is no a_empt at sliding. Thus, no fric>on and no mo>on. NO FRICTION Force F aKempts sliding but is balanced by the fric>onal force. No mo>on. STATIC FRICTION Force F is now stronger but is s>ll balanced by the fric>onal force. No mo>on. LARGER STATIC FRICTION Force F is now even stronger but is s>ll balanced by the fric>onal force. No mo>on. EVEN LARGER STATIC FRICTION Finally, the applied force has overwhelmed the sta>c fric>onal force. Block slides and accelerates. WEAK KINETIC FRICTION To maintain the speed, weaken force F to match the weak fric>onal force. SAME WEAK KINETIC FRICTION Sta>c fric>onal force can only match growing applied force. Kine>c fric>onal force has only one value (no matching). fs is the sta>c fric>onal force fk is the kine>c fric>onal force Sample Problem, fric>on applied at an angle Hint: Develop an expression for accelera+on, then use calculus to find its maximum. Sample Problem, fric>on applied at an angle 6.4: The drag force and terminal speed
When there is a rela>ve velocity between a fluid and a body (either because the body moves through the fluid or because the fluid moves past the body), the body experiences a drag force, D, that opposes the rela>ve mo>on and points in the direc>on in which the fluid flows rela>ve to the body. 6.4: Drag force and terminal speed
For cases in which air is the fluid,
and the body is blunt (like a
baseball) rather than slender (like a
javelin), and the relative motion is
fast enough so that the air becomes
turbulent (breaks up into swirls)
behind the body,
where ρ is the air density (mass per volume), A is the effec5ve cross-­‐
sec5onal area of the body (the area of a cross sec>on taken perpendicular to the velocity), and C is the drag coefficient . Terminal SPEED
When a blunt body falls from rest through air, the drag force is directed upward; its magnitude gradually increases from zero as the speed of the body increases. Eventually, a = 0, and the body then falls at a constant speed, called the terminal speed vt . 6.4: Drag force and terminal speed
For cases in which air is the fluid, and the body is blunt (like a baseball) rather than slender (like a javelin), and the rela>ve mo>on is fast enough so that the air becomes turbulent (breaks up into swirls) behind the body, where ρ is the air density (mass per volume), A is the effec5ve cross-­‐
sec5onal area of the body (the area of a cross sec>on taken perpendicular to the velocity), and C is the drag coefficient . When a blunt body falls from rest through air, the drag force is directed upward; its magnitude gradually increases from zero as the speed of the body increases. From Newton’s second law along y axis where m is the mass of the body. Eventually, a = 0, and the body then falls at a constant speed, called the terminal speed vt . 6.4: Drag force and terminal speed
Some typical values of terminal speed
SAMPLE PROBLEM: Terminal Speed A raindrop with radius R = 1.5 mm falls from a cloud that is at height h = 1200 m above the ground. The drag coefficient C for the drop is 0.60. Assume that the drop is spherical throughout its fall. (a)  What is the terminal speed of the drop? (b) What would be the drop’s speed just before impact if there were no drag force? Sample problem, terminal speed
FORCES IN CIRCULAR MOTION
6.5: Uniform circular motion
Example of a hockey puck: Fig. 6-­‐8 An overhead view of a hockey puck moving with constant speed v in a circular path of radius R on a horizontal fric>onless surface. The centripetal force on the puck is T, the pull from the string, directed inward along the radial axis r extending through the puck. Nega>ve LiN Derived Equa>ons for (1)  banking with no fric>on, and (2)  fric>on with no banking (a flat curve) BANKED CURVE (with/without fric>on)