Recall: Applying Newton’s Laws
► Assumptions
Objects behave as particles
► can ignore rotational motion (for now)
Masses of strings or ropes are negligible
Interested only in the forces acting on the
object
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Applying Newton's Laws: Solving Problems Involving Forces and Friction, Slides of Physics
A recall of newton's laws, examples of their application, and solutions to related problems. Topics covered include applying the laws to various situations, making sketches and free body diagrams, resolving forces into components, and using equations to find unknowns. Problems involve calculating coefficients of friction, forces, acceleration, and terminal speeds.
Typology: Slides
2012/2013
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Recall: Applying Newton’s Laws
► Assumptions
Objects behave as particles
►can ignore rotational motion (for now)
Masses of strings or ropes are negligible
Interested only in the forces acting on the
object
Recall: Applying Newton’s Laws
► Make a sketch of the situation described in the
problem, introduce a coordinate frame
► Draw a free body diagram for the isolated object
under consideration and label all the forces acting
on it
► Resolve the forces into x- and y-components,
using a convenient coordinate system
► Apply equations, keeping track of signs
► Solve the resulting equations
6 Example: A box full of books rests on a wooden floor. The normal force the floor exerts on the box is 250 N. (a) You push horizontally on the box with a force of 120 N, but it refuses to budge. What can you say about the coefficient of friction between the box and the floor? FBD for box: N w x y F f s
( 1 ) F
∑ (^) y
= N − w = 0
( 2 ) F
∑ (^) x
= F − f
s
Apply Newton’s 2 nd Law From (2): μ s
0.48. s s s s N f f ≤ μ N ⇒ ≤ μ s s s N N N F N f F = f ⇒ = = = 0. 48 < μ 250 120 The box refuses to budge, s s N f ⇒ < μ
7 N w x y F
f
s,max (b) If you must push horizontally on the box with 150 N force to start it sliding, what is the coefficient of static friction? ( 2 ) 0 ( 1 ) 0 , max = − = = − = ∑ ∑ x s y F F f F N w Apply Newton’s 2 nd Law s s s s N f f ≤ μ N ⇒ ≤ μ The box starts sliding, the friction force is fs,ma x here From (2): (^0). 60 250 ,max^150 , max s = ⇒ = = = = N N N F N f F f s s
μ s = 0.60.
9 Example: A box slides across a rough surface. If the coefficient of kinetic friction is 0.3, what is the acceleration of the box? If the initial speed of the box is 10.0 m/s, how long does it take for the box to come to rest? F k w N x y FBD for box:
∑ ∑
F N w
F F ma
y x k Apply Newton’s 2 nd Law:
Do you remember kinematics?
10 N w N w mg F ma k − = ∴ = = − = 0
F N mg ma
k k k
(1) (2) From (1):
2 2 a = − g = − 0. 3 9. 8 m/s = − 2. 94 m/s k μ Solving for a: F k w N x y
Air Resistance and Terminal
Speed
► Another type of friction is air resistance
► Air resistance is proportional to the speed of
the object
► When the upward force of air resistance
equals the downward force of gravity, the
net force on the object is zero
► The constant speed of the object is the
terminal speed
13 Imagine that a stone is dropped from the edge of a cliff. If
air resistance cannot be ignored, the FBD for the stone is:
Apply Newton’s Second Law F F w ma y d = − = ∑ x y w Fd Where F d is the magnitude of the drag force on the stone. This force is directed opposite the object’s velocity.
Air Resistance
15 Example: A paratrooper with a fully loaded pack has a mass of 120 kg. The force due to air resistance has a magnitude of F d = bv 2 , where b = 0.14 N s 2 /m 2 . (a) If he/she falls with a speed of 64 m/s, what is the force of air resistance? ( 0. 14 Ns /m )( 64 m/s) 570 N 2 2 2 2
F = bv = =
d
16 (b) What is the paratrooper’s acceleration? Apply Newton’s Second Law and solve for a. 2 = − 5. 1 m/s − = = − = ∑ m F mg a F F w ma d y d x y w Fd FBD: (c) What is the paratrooper’s terminal speed? 92 m/s 0 0 2 = = − = ∑ = − =^ = b mg v bv mg F F w ma t t y d Example continued:
18 An ideal cord has zero mass, does not stretch, and the tension is the same throughout the cord. Tension is the force transmitted through a “rope” from one end to the other.
Tension
19 Example: Find the tension in the cord connecting the two blocks as shown. A force of 10.0 N is applied to the right on block 1. Assume a frictionless surface. The masses are m 1 = 3.00 kg and m 2 = 1.00 kg. F block 2 block 1 Assume that the rope stays taut so that both blocks have the same acceleration.
21 F T m a 1 − = T m a 2 = These two equations contain the
unknowns: a and T.
To solve for T, a must be eliminated. Solve for a in (2) and substitute in (1). (1) (2)
- 5 N 1 kg 3 kg 1 10 N 1 1 2 1 2 1 2 1 2 1 1 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛
∴ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛
- = + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = = m m F T T m m T m T F m m T F T ma m Example continued:
Connected Objects
► Apply Newton’s Laws separately to each object ► The acceleration of both objects will be the same ► The tension is the same in each diagram ► Solve the simultaneous equations This example leads to the following observations/suggestions: