Download Physics

AP Physics 3: Forces—Dynamics
A.
B.
Newton’s Laws of Motion
1. Aristotle's view was that an object's natural state is
rest and it takes a force (push or pull) to keep an
object moving
2. Galileo's view was that an object's natural state was
unchanged motion, either at rest or at a constant
speed in a straight line (Law of Inertia)
3. Newton synthesized causes of motion into three laws
a. First Law (Galileo's law of Inertia): object remains
at rest or uniform velocity in a straight line as long
as no net force (Fnet) acts on it
b. Second Law: (Fnet = ma)
1. measured in newtons: 1 N = 1 kg•m/s
2. Fnet  and v : v increases
Fnet  and v : v decreases
Fnet  and v : v turns in a circle
3. Ft = mv
Steps
Algebra
F = ma
start with
multiply both sides by t
Ft = mat
substitute v/t for a
Ft = m(v/t)t = mv
a. mv is Newton's "quantity of motion"
b. now called momentum, p = mv (kg•m/s)
c. Third Law: action force on A generates an equal
but opposite reaction force on B (FA = -FB)
d. four important concepts
1. force can act on contact (collision) or at a
distance (gravity)
2. usually multiple forces act on an object  the
vector sum of all forces = Fnet
3. mass is measured in terms of Newton's laws
a. inertial mass = object's resistance to
change in motion (first law)
b. gravitational mass = gravity's affect on
an object (second law)
4. third law forces are equal and opposite, but
don't cancel each other out because they act
on different objects, which can cause either
or both objects to accelerate.
Types of Forces
1. push or pull (Fp)
a. measured using a spring scale (force increases
linearly as distance that a spring is stretched x
increases)
1. spring force, Fs = kx
2. k is the spring constant
b. tension (Ft or T) can be used instead of Fp
2. weight (Fg or W) is the force of attraction between the
object and the Earth—gravity, Fg = mg
a. g = 9.80 m/s2 (negative sign is not included)
b. directed down to the Earth’s center
3. normal force (Fn or N) is the force that the surface
exerts on an object to support its weight
a. perpendicular away from the surface
b. not calculated in isolation, but is determined by
other perpendicular forces so that F = 0
4. friction (Ff) is parallel to surface and opposes motion
a. when moving: Ff kFn
(k = kinetic coefficient of friction)
b. when stationary: Ff is part of F|| = 0, but cannot
exceed Ff  sFn (s = static coefficient of friction)
5. free-body diagram
a. diagram shows all forces acting on the system
b. Fp/Ft: along direction of push or pull
c. Ff: opposes motion and is || to surface
d. Fg—toward Earth's center
e. Fn— to surface
Name __________________________
C.
Force Problems—Dynamics
1. general set-up

draw a free body diagram

resolve forces into || and  components to motion

assign positive directions
o for perpendicular forces, up is positive
o for parallel forces, direction of velocity is positive

two equations
o  F = 0
o F|| = ma

calculate d, v and t using kinematics
2. horizontal surface
Fp
Fn
Fp- = Fpsin

Ff 
Fp-|| = Fpcos



3.
Fn = Fg – Fp-
Fg = mg
Ff = Fn
F|| = Fp-|| – Ff = ma (m is everything that moves)
incline (moving up)
Fn
Fp

Ff
 Fg= mg
Fg- = Fgcos
Fg-|| = Fgsin

Fn = Fg-


Ff = Fn

F|| = Fp – Ff – Fg-|| = ma (m is everything that moves)
5. internal tension
treat the system as one object
Ff
Ft-1 = Ft-2
m1


D.
Fp
m2
Ft-1 = Ft-2 (third law)  cancel out
F|| = Fp – Ff = ma (m is everything that moves)
a = (Fp – Ff)/(m1 + m2)
isolate one part

Fp – Ft-2 = m2a (a is for the whole system)

Ft-1 – Ff = m1a (a is for the whole system)
6. pulleys
treat the system as one object (m2 > m1)

T1 = T2 (third law)  cancel out
+a 
 +a

F|| = W2 – W1 = ma

a = (m2 – m1)g/(m2 + m1)
T1
T2
isolate one part
m1
m2

W2 – T2 = m2a
W1
W2

T1 – W1 = m1a
7. vertical acceleration

Fn (platform) or Fp (rope, rocket) generates acceleration

Fn/p – Fg = ma
o +a, apparent weight > normal
o –a, apparent weight < normal
o –a = g (weightless)
Force Problems—Statics
1. one unknown

draw a free body diagram

resolve forces into || and  components to motion

assign positive directions

two equations
o  F = 0
o F|| = 0

calculate d, v and t using kinematics
2.
two unknowns
 L
R
TL




TR
Fg = mg
Fy = TLsin(180 – L) + TRsin(R) + Fgsin(-90) = 0
Fx = TLcos(180 – L) + TRcos(R) + Fgcos(-90) = 0
solve for TL in terms of TR in the second equation and
then substitute into the first equation
special case (two of three forces are )
L
TL
Fg


sinL = Fg/TL
tanL = Fg/TR
TL
L
TR
TR
Fg = mg
A.
Newton's Laws of Motion
Questions 1-11 Briefly explain your answer.
Atwood Machine Lab
1. A book is lying at rest on a table because
Time how long it takes the upper weight with additional mass
(A) there are no forces acting on the book.
mp to descend the measured distance, graph the data,
(B) the forces cancel each other out.
calculate the acceleration using kinematics and Newton's law
B—Fg and Fn cancel each other out
and compare the two accelerations.
a. Collect the following data.
2. A hockey puck slides on ice at constant velocity. What is
Descending Distance
the direction of the net force on the puck?
d (m)
(A) forward
(B) backward
(C) no net force
Time
(s)
mp
Fp
Exp.
(g)
(N)
trial 1 trial 2 trial 3 Average
C—Fnet = ma = m(0) = 0
1
15
0.147
2
16
0.157
3
17
0.167
4
18
0.176
5
19
0.186
6
20
0.196
You kick a smooth flat stone out on a frozen pond. The
stone slides, slows down and eventually stops. What is the
direction of the net force on the stone?
(A) forward
(B) backward
(C) no net force
7
21
0.206
B—Friction slows the stone  force is backward
3.
B—Friction slows the book  force is backward
4.
b.
Calculate the acceleration using kinematics.
Calculation
Formula
1
2
3
4
5
6
5.
7
c.
Consider a cart on a horizontal frictionless table. Once the
cart has been given a push and released, what will happen
to the speed of the cart?
(A) slow down (B) constant
(C) speed up
B—Fnet = 0  no acceleration and the speed stays the
same
d = ½at2
a
You put your book on the bus seat next to you. When the
bus stops suddenly, the book slides forward a short
distance. What is the direction of the net force on the book?
(A) forward
(B) backward
(C) no net force
Graph the Fp vs. a and draw a best fit line (excluding
the origin).
Fp (N)
0.20
Questions 6-8 From rest, you step on the gas of a Ferrari,
providing a force F for time t, resulting in final speed v and
distance traveled d.
6. How much time would be needed to reach speed v using ½F?
(A) 4t
(B) 2t
(C) ½t
(D) ¼t
B—Ft = mv, but mv is constant  Ft is constant
Ft = ½Ft  t = 2t
0.16
0.12
7.
0.08
How much time would be needed to reach a speed of 2v?
(A) 4t
(B) 2t
(C) ½t
(D) ¼t
B—Ft = mv, double v doubles t
0.04
8.
0
0.25
0.35
0.45
a (m/s2)
d. Determine the y-intercept. This is Ff = ________
e. Calculate the net force, total mass and acceleration.
Experiment
1
2
3
4
5
6
7
Fnet
0.05
Fp - Ff
mtotal .4 + mp
0.15
How far would the car travel if t is doubled?
(A) 4d
(B) 2d
(C) ½d
(D) ¼d
A—Ft = mv, double t doubles v, but v2 = 2ad 
double t quadruples d
9.
A force F acts on a mass M1, giving acceleration a1. The
same force acts on a different mass M2, giving acceleration
a2 = 2a1. If M1 and M2 are glued together and the same force
F acts on this combination, what is the resulting acceleration?
(A) 3/2a1
(B) 1/2a1
(C) 2/3a1
(D) 1/3a1
C—F = ma: a2 = 2a1 m2 = ½m1 and mtot = 3/2m1
 a = 2/3a1
Questions 10-11 David Scott (Apollo 15) dropped a feather and
hammer on the Moon from the same height and they
reached the ground at the same time. Which object had a
f. Calculate the percent difference in a (from parts b and e)
greater acceleration due to gravity?
for each experiment and average over all.
(A) feather
(B) hammer
(C) tie
Experiment
1
2
3
4
5
6
7
C—Both fell at the same rate  same acceleration.
%
a
Fnet/mtot
Average
10. Which had the greater force of gravity?
(A) feather
(B) hammer
(C) tie
B—Both have equal accelerations, but the hammer
has greater mass (Fg = ma)
11. David Scott throws a ball on Earth. On the Moon, he
throws the same ball with the same force. The
acceleration of the ball on the Moon compared to Earth is
(A) more
(B) less
(C) the same
C—F = ma: with the same F and m, a will be the same
12. What is the direction of the net force for each situation?
B. Types of Forces
Static Friction Lab
Part 1: pull on the wood block with weights on top from the
same initial spot near the top of the incline using the spring
scale until the block just begins to move, record the
greatest force reading on the spring scale (Fs), graph the
data and use the slope to determine s.
a. Collect the following data.
Mass of wood block, M1 (kg)
Weights, M2 (kg)
0
0.10 0.20 0.30 0.40 0.50
northward
car is accelerating northward
Fn Fn = (M1 + M2)g
bowling ball rolling straight at constant speed
zero
thrown rock reaches its highest point
downward
rock resting on the ground
zero
13. Determine the net force and acceleration on the 10-kg box
for each situation.
30 N
30 N
Fnet = -30 N + 30 N = 0 N
Fnet = ma
0 N = (10 kg)a  a = 0 m/s2
30 N
30 N
50 N
20 N
F f F f = Fs
b.
Graph the Ff vs. Fn and draw a best fit line.
Ff
2.5 N
2.0 N
1.5 N
1.0 N
Fnet = -30 N + 50 N = 20 N
0.5 N
Fnet = ma
20 N = (10 kg)a  a = 2 m/s2
0.0 N
Fnet = -30 N + 20 N = -10 N
Fnet = ma
-10 N = (10 kg)a  a = -1m/s2
14. State whether the pair of equal forces are third law forces
or first law forces.
a. Force between two ice skaters pushing on each other.
Third Law
b. Tension in a cord equals the weight of a hanging mass.
First Law
15. Two 1-kg weights are suspended from a frictionless pulley
(an Atwood machine). An additional 0.0500 kg is added to
the upper weight and allowed to fall freely.
a. What is the sum of forces acting on the system?
c.
1.0 N
3.0 N
5.0 N
Fn
Use the slope of the line to determine s.
Part 2: measure the angle of the incline, which is needed
for the wood block to just begin moving (angle of repose).
d. Collect the following data
Trial
1
2
3
Average

tan
e.
Label all the forces acting on the block in terms of Fg.
Fn = Fgcos
Ff = Fgsin
 Fg = Fgcos
Fnet = mg = (0.0500 kg)(10 m/s2) = 0.500 N
b.
Fg
What is the total mass of all moving parts?
Fg-|| = Fgsin
f.
Determine an
expression for  in terms of .

mtotal = 1 kg + 1 kg + 0.05 kg + 2.05 kg
c.
What is the theoretical acceleration?
Fnet = ma
0.500 N = (2.05 kg)a  a = 0.244 m/s2
d.
e.
 = Ff/Fn
 = Fgsin/Fgcos = tan
g.
The system takes 2.90 s to move 1.00 m. What is the
actual acceleration?
d = ½at
1.00 m = ½a(2.90 s)2  a = 0.238 m/s2
What is the percent difference between theoretical and
actual accelerations?
%  = 100(0.244 – 0.238)/0.244 = 2 %
7.0 N
h.
Calculate the percent difference between  (part c)
and tan ( part d).
Does placing the block on its side significantly change
the angle of repose? Find out.
Trial
1
2
3
Average 
Angle of repose
Affect on 
Fg = mgmoon = (75 kg)(1.6 m/s2) = 120 N
Questions 16-22 Briefly explain your answer.
16. Consider two identical blocks; A resting on a horizontal
surface and B resting on an incline. Which block has a
greater normal force?
(A) NA > NB
(B) NA < NB
(C) NA = NB
A—NA = Fg, but NB = Fgcos, which is less than Fg
17. When you walk forward, what is the direction of the net
force you exert on the ground?
(A) forward
(B) backward
(C) downward
B—Newton's third law reaction force generates the
forward force
Questions 18-21 Below you see two cases: a physics student
pushing (a) or pulling (b) a sled with a force F, which is
applied at an angle .
18. Which is true of the normal force N on the sled?
(A) Na > Nb
(B) Na < Nb
(C) Na = Nb
A—Na = mg + Fsin and Nb = mg – Fsin  Na > Nb
19. Which is true of the force of friction Ff on the sled?
(A) Ffa > Ffb
(B) Ffa < Ffb
(C) Ffa = Ffb
A—Ffa = Na and Ffb = Nb, but Na > Nb  Ffa > Ffb
20. Which is true of the force F need for constant speed?
(A) Fa > Fb
(B) Fa < Fb
(C) Fa = Fb
A—since Fa = Ffa and Fb = Ffb, but Ffa > Ffb Fa > Fb
21. Which is true of the acceleration if the force F is the same?
(A) aa > ab
(B) aa < ab
(C) aa = ab
B—F – Ff = ma, since Ffa > Ffb, then less force is available
for acceleration  aa < ab
22. A box generates a maximum force of static friction equal to
50 N. A horizontal force of 30 N is directed to the right.
What direction will the box move?
(A) right
(B) left
(C) the box doesn't move
25. Determine the normal force on 10 kg block that is
a. resting on a horizontal surface.
Fn = Fg = mg = (10 kg)(10 m/s2) = 100 N
b.
Fn = Fg + Fp = 100 N + 25 N = 125 N
c.
resting on a horizontal surface with a force of 25 N
pulling up on the block.
Fn = Fg – Fp = 100 N – 25 N = 75 N
resting on a 30o incline.
d.
Fn = Fgcos= (100 N)cos30 = 87 N
26. A 10-kg block rests on a surface. The coefficients of
friction are s = 0.3, k = 0.15.
a. What is the maximum F|| that can act on the stationary
block?
F = Ff = sFn = (0.3)(10 kg)(10 m/s2) = 30 N
b.
What is the force of friction when F|| = 10 N?
Ff = 10 N
c.
What is the force of friction when F|| = 40 N?
Ff = kFn = (0.15)(100 N) = 15 N
C.
Force Problems—Dynamics
Kinetic Friction Lab
Time how long it takes the wood block to rise to the top of
the incline (if it takes more than 1.5 s, change the angle of
the incline)
a. Collect the following data.
Hanging weight M1 Block mass M2 distance d
Angle 
(kg)
(kg)
(m)
Time
(s)
C—Friction can only resist motion, not cause it, since
the horizontal force < force of friction, then no motion
23. A 100-N force Fp pulls on a 10 kg block at an angle of 30o
from horizontal along a frictionless surface.
a. What is the perpendicular component of Fp?
resting on a horizontal surface with a force of 25 N
pressing down on the block.
b.
a
Trial 1
Trial 2
Calculate the following from the data.
Formula
Calculation
d = ½at2
Fp- = Fpsin30 = (100 N)sin30 = 50 N
b.
What is the parallel component of Fp?
Fp- = Fpcos30 = (100 N)cos30 = 87 N
c.
m = M1 + M2
Fg-M1 Fg-M1 = M1g
What is the acceleration of the block?
Fg-M2|| Fg-M2|| = M2gsin
F|| = ma
87 N = (10 kg)a  a = 8.7
m/s2
24. What is the force of gravity (weight) of a 75 kg person?
a. On earth (g = 9.8 m/s2)
Fg = mgearth = (75 kg)(9.8 m/s2) = 735 N
b.
mtot
On the moon (g = 1.6 m/s2)
Trial 3
Ff
Fg-M1 – Fp – Fg-M2|| = mtota
Fn
Fn = FgM2cos
tav
C—F|| = mgsin – mgcos= ma: m cancels out of the
equation  m doesn't affect the situation
Ff = Fn

Questions 27-37 Briefly explain your answer.
27. Two blocks of masses 2m and m are in contact on a
horizontal frictionless surface. If a force F is applied to
mass 2m, what is the force on mass m?
2m
F
m
(A) 2 F
(B) F
(C) 1/2F
(D) 1/3F
D—Force is proportional to mass when acceleration is
the same, since m = 1/3 total mass then force = 1/3F
28. Three blocks of mass 3m, 2m and m are connected by
strings and pulled with constant acceleration a. What is the
relationship between the tensions in each string?
3m
T3
T2
2m
T1
m
(A) T1 > T2 > T3 (B) T3 > T2 > T1 (C) T1 = T2 = T3
A—T1 pulls all the mass, T2 pulls 5/6 of the mass and T3
pulls ½ of the mass, T1 > T2 > T3
29. You tie a rope to a tree and you pull on the rope with a
force of 100 N. What is the tension in the rope?
(A) 0 N
(B) 50 N
(C) 100 N (D) 200 N
C—Newton's third law of equal and opposite forces
30. Two tug-of-war opponents each pull with a force of 100 N on
opposite ends of a rope. What is the tension in the rope?
(A) 0 N
(B) 50 N
(C) 100 N (D) 200 N
C—Whether tied to a tree or pulled with equal force, the
tension is the same, 100 N
31. You tie a rope to a tree and you and your friend each pull
on the free end of the rope with forces of 100 N. What is
the tension in the rope?
(A) 0 N
(B) 50 N
(C) 100 N (D) 200 N
D—The combined forces on one end will generate a
reaction force equal to 200 N
32. In which case does the acceleration the greatest; in case 1,
where a 10-kg hanging mass generates a downward force
of 100 N or case 2 where a hand generates a downward
force of 100 N?
50 kg
(A) case 1
(B) case 2
(C) both cases are the same
100 N
B—F = ma, where m = 50 kg + 10 kg for case 1 and 50 kg
for case 2  case 2 where the mass is less
33. A box sits on a flat board. You lift one end of the board,
making an angle with the floor. As you increase the angle,
the box will eventually begin to slide down. Why?
(A) component of the gravity parallel to the plane increases
(B) normal force exerted by the board decreases
(C) both (A) and (B)
C—F|| = Fgsin – Fgcos: as  increases the first term
increases and the second term decreases
34. A block of mass m slides down a rough incline at constant
speed. If a similar block of mass ½m were placed on the
same incline, its speed would
(A) be zero
(B) increase
(C) be constant
Questions 35-37 A block of mass m rests on the floor of an
elevator. Use the following options.
(A) Fn > Fg (B) Fn = Fg (C) Fn < Fg (D) Fn = 0
35. Which is true if elevator is at constant velocity downward?
B—Since acceleration is zero, Fnet = Fn – Fg = 0
 Fn = F g
c.
d.
37. Which is true if the elevator is allowed to fall freely?
D—Since acceleration is equal to -g, Fnet = Fn – Fg = -mg
Fn = Fg – mg = 0
38. A stationary 5-kg box is pulled upward by a 70-N force, Fp.
a. Draw all the forces acting on the box.
Fp = 70 N
e.
force of friction Ff. (k = 0.10)
Ff = Fn
Ff = 0.10(690 N) = 69 N
f.
overall force acting on the skier parallel to the surface.
Fnet = Fg-|| – Ff
Fnet = 400 N – 69 N = 330 N
g.
skier’s acceleration.
Fnet = ma
330 N = (80 kg)a  a = 4.1 m/s2
41. An elevator (m1 = 1150 kg) is suspended over a pulley
which is connected to a counterweight (m2 = 1000 kg).
a. Draw all forces acting on the elevator and
counterweight.
Determine the
Fg = 50 N
b. overall force acting on the box.
Fnet = Fp – Fg
Fnet = 70 N – (5 kg)(10 m/s2) = 20 N
c.
normal force Fn.
Fn = Fg- = 690 N
36. Which is true if the elevator is accelerating upward?
A—Since acceleration is upward, Fnet = Fn – Fg > 0
Fn > Fg
perpendicular component Fg-
Fg- = Fgcos
Fg- = (80 kg)(10 m/s2)cos30 = 690 N
acceleration of the box.
Fnet = ma
20 N = (5 kg)a  a = 4 m/s2
d.
Ft-1
vertical velocity of the box after 2 s of acceleration.
vt = vo + at
v = 0 + (4 m/s2)(2 s) = 8 m/s upward
Ft-2
39. A 0.5-kg hockey puck traveling at 10 m/s on a horizontal
surface slows to 2 m/s over a distance of 80 m.
a. Draw all the forces acting on the hockey puck
Fn
Ff
Determine the
Fg
b. acceleration of the puck (use kinematics).
vt2 = vo2 + 2ad
(2 m/s)2 = (10 m/s)2 + 2a(80 m)  a = -0.6 m/s2
c.
force of friction acting on the puck.
Ff = ma
Ff = (0.5 kg)(-0.6 m/s2) = -0.3 N
d.
Determine the
Fg-2 = 100,000 N
b. overall force acting on the system.
Fnet = Fg-1 – Ft-1 + Ft-2 – Fg-2 = 11500 N – 10000 N = 1500 N
c.
acceleration.
Fnet = ma
1500 N = (m1 + m2)a = (2150 kg)a  a = 0.70 m/s2
normal force acting on the puck.
Fn = Fg = mg
Fn = (0.5 kg)(10 m/s2) = 5 N
e.
Fg-1 = 11500 N
d.
tension in the cable above the elevator.
Fnet = Fg-1 – Ft-1 = m1a
11500 N – Ft-1 = (1150 kg)(0.70 m/s2)  Ft-1 = 10700 N
coefficient of kinetic friction.
e.
Ff = Fn
0.3 N = (5 N)   = 0.06
tension in the cable above the counterweight.
Fnet = Ft-2 – Fg-2 = m2a
Ft-2 – 10000 N = (1000 kg)(0.70 m/s2)  Ft-2 = 10700 N
40. An 80-kg skier descends a 30o slope.
a. Draw all forces acting on the skier.
Fn
Ff
f.
time it takes to travel 3 m starting from rest.
d = vot + ½at2
3 m = 0 + ½(0.70 m/s2)t2  t = 2.9 s
g.
speed of the elevator after traveling 3 m from rest.
v2
= vo2 + 2ad = 0 + 2(0.70 m/s2)(3 m) = 4.2 m2/s2
v = 2.0 m/s
h.
Fg
Determine the
b. parallel component of the force of gravity Fg-||.
Fg-|| = Fgsin
Fnet = (80 kg)(10 m/s2)sin30 = 400 N
the apparent weight of a 50-kg elevator passenger.
Fnet = Fg – Fn = ma
500 N – Fn = (50 kg)(0.70)  Fn = 465 N
42. A 25-kg block is pulled vertically upward by a force of 325 N.
a. Draw all the forces acting on the block to the right.
Determine the
Fp = 325 N
b. overall force acting on the block.
45. A 10-kg block slides down a 37o incline.
a. Draw all forces acting on the block. Ff
Fn
Fnet = Fp – Fg = 325 N – 250 = 75 N
 Fg
Fg
c.
block's acceleration.
Fnet = ma
75 N = (25 kg)a  a = 3 m/s2
d.
block's vertical velocity after 2 s of motion. Fg = 250 N
vt = vo + at = 0 + (3 m/s2)(2 s) = 6 m/s
43. Object A (mA = 10 kg) at rest on a table ( = 0.30) is
attached to B (mB = 5 kg).
a. Draw all forces acting on A and B.
Fn
Ff
Ft-B
Fg-A
Fg-B
overall || force on the system.
Fnet = Fg-B – Ff
Fnet = (5 kg)(10 m/s2) – 30 N = 20 N
d.
acceleration of the system.
Fnet = (mA + mB)a
20 N = (10 kg + 5 kg)a  a = 1.3 m/s2
e.
c.
perpendicular component Fg-
Fg- = Fgcos
Fg- = (10 kg)(10 m/s2)cos37 = 80 N
normal force Fn.
Fn = Fg- = 80 N
e.
Ff = Fn = mAg
Ff = (0.30)(10 kg)(10 m/s2) = 30 N
c.
Fg-|| = Fgsin
Fg-|| = (10 kg)(10 m/s2)sin37 = 60 N
d.
Ft-A
Determine the
b. force of friction on A.
Fg-||

Determine the
b. parallel component of the force of gravity Fg-||.
force of friction Ff. (k = 0.25)
Ff = Fn
Ff = 0.25(80 N) = 20 N
f.
overall force acting on the block.
Fnet = Fg-|| – Ff
Fnet = 60 N – 20 N = 40 N
g.
block's acceleration.
Fnet = ma
40 N = (10 kg)a  a = 4 m/s2
46. A 1.5-kg weight is suspended over a pulley which is
connected to a 1-kg weight
a. Draw all forces acting on the 1.5-kg and 1-kg weights.
tension in the rope.
Fnet = Fg-B – Ft = mBa
(5 kg)(10 m/s2) – Ft = (5 kg)(1.3 m/s2)  Ft = 44 N
f.
Ft-1
velocity of B after it has descended 0.5 m.
Ft-1.5
vt2 = vo2 + 2ad
vt2 = 0 + 2(1.3 m/s2)(0.5 m)  vt = 1.1 m/s
44. A 0.50-kg hockey puck traveling at 15 m/s on a horizontal
surface slows to a stop in 90 m.
a. Draw all the forces acting on the hockey puck
Fn
Ff
Determine the
Fg
b. puck's acceleration (use kinematics).
vt2 = vo2 + 2ad
(0 m/s)2 = (15 m/s)2 + 2a(90 m)  a = -1.25 m/s2
c.
force of friction.
Ff = ma
Ff = (0.50 kg)(-1.25 m/s2) = -0.625 N
d.
normal force.
Fn = Fg = mg
Fn = (0.50 kg)(10 m/s2) = 5 N
e.
coefficient of kinetic friction.
Ff = Fn
0.625 N = (5 N)   = 0.125
Fg-1
Fg-1.5
Determine the
b. overall force acting on the system.
Fnet = Fg-1.5 – Ft-1.5 + Ft-1 – Fg-1
Fnet = (m1.5 – m1)g = (1.5 kg – 1 kg)(10 m/s2) = 5 N
c.
acceleration.
Fnet = ma
5 N = (2.5 kg)a  a = 2 m/s2
d.
tension in the cable above the 1.5-kg weight.
Fnet = Fg-1.5 – Ft-1.5 = m1.5a
15 N – FT-1.5 = (1.5 kg)(2 m/s2)  Ft-1.5 = 12 N
e.
tension in the cable above the 1-kg weight.
Fnet = Ft-1 – Fg-1 = m1a
Ft-1 – 10 N = (1 kg)(2 m/s2)  Ft-1 = 12 N
f.
time it takes to travel 2.25 m starting from rest.
d = vot + ½at2
2.25 m = 0 + ½(2 m/s2)t2  t = 1.5 s
D.
Force Problems—Statics
Force Table Lab
Determine the missing force and angle using spring scales
and compare the results to the calculated values.
a. Collect the following data.
Scale A
FA

Experiment
b.
Scale B
FB

48. A 50-kg box is anchored to the ceiling by two cords.
45o
30o
Scale C
FC

1
0o
1.0 N
90o
1.0 N
2
20o
1.0 N
80o
0.5 N
a.
3
230o
2.0 N
0o
1.3 N
TLcos135 + TRcos30 = 0
TL(-0.707) + TR(0.866) = 0  TR = TL(0.816)
Calculate scale C force and angle from scale A and
scale B values.
Experiment
1
2
3
Formula
Calculation
FAx Ax = Acos
Use Fx = 0 to write an expression for the tension in
the right cable in terms of the tension in the left cable.
b. Use Fy = 0 to solve for the tension in the left cable.
TLsin135 + TRsin30 + mgsin-90 = 0
TL(0.707) + TL(0.816)(0.500) + (30 kg)(10 m/s2)(-1) = 0
TL(1.12) = 300 N  TL = 269 N
c. Substitute the left tension into the expression from (a).
TR = TL(0.816) = (269 N)(0.816) = 220 N
FBx Bx = Bcos
49. A 50-kg box is anchored to the ceiling and wall by cords.
FCx Ax + Bx + Cx = 0
60o
FAy Ay = Asin
TL
FBy By = Bsin
TR
FCy Ay + By + Cy = 0
a.
FC
C=
(Cx2
Label each force in the vector sum diagram.
2 )½
+ Cy
Fg

tan = Cy/Cx
c.
Calculate the percent difference between the measured
values of (part a) and theoretical values of (part b).
Experiment
1
2
3
Formula
Calculation
b.
% =100||/C

47. The 100-N block is stationary and s = 0.40.
60o
TR
Calculate the tension in the ceiling cord (TL).
sin60 = Fg/TL = 500 N/TL  TL = 577 N
c.
Fc
Calculate the tension in the wall cord (TR).
tan60 = Fg/TR = 500 N/TR  TR = 289 N
50. A 50-kg box is anchored to the ceiling by two cords, which
form a 90o angle between them.
53o
TL
a.
What is the minimum weight W?
W + Ff = 75 N
W = 75 – Fg = 75 – (0.40)(100) = 35 N
b.
TL
b.
37o
TR
Calculate TL.
sin53 = TL/Fg = TL/500 N  TL = 400 N
What is the maximum weight W?
W = 75 + Ff
W = 75 + 40 = 115 N
c.
Calculate TR.
cos53 = TR/Fg = TR/500 N  TR = 300 N
Practice Multiple Choice (No calculator)
Briefly explain why the answer is correct in the space provided.
1. A 6-N force and an 8-N force act on a point. The resulting
force is 10 N. What is the angle between the two forces?
(A) 0o
(B) 30o
(C) 45o
(D) 90o
10. Two 0.60-kg objects are connected by a thread that
passes over a light, frictionless pulley. The objects are
initially held at rest. A 0.30-kg mass is added on top of one
of the 0.60-kg objects and the objects are released.
D—The vector sum forms a 6-8-10 (3-4-5) triangle,
which is a right triangle, where the 3-4 sides make 90o
2.
A 60-kg physics student would weigh 1560 N on the
surface of planet X. What is the acceleration due to gravity
on the surface of planet X?
(A) 2 m/s2 (B) 10 m/s2 (C) 15 m/s2 (D) 26 m/s2
D—Fg = mg
1560 N = (60 kg)g g = 26 m/s2
3.
A 400-N girl standing on a dock exerts a force of 100 N on
a 10,000-N sailboat as she pushes it away from the dock.
How much force does the sailboat exert on the girl?
(A) 25 N
(B) 400 N (C) 100 N (D) 10000 N
The acceleration of the system of objects is most nearly
(A) 10 m/s2 (B) 6 m/s2 (C) 3 m/s2 (D) 2 m/s2
C—Fnet = ma (Fnet = extra weight and m = total mass)
(0.30 kg)(10 m/s2) = (.30 + .60 + .60 kg)a  a = 2 m/s2
11. The diagram shows a 4.0-kg object accelerating at 10 m/s2
on a rough horizontal surface.
C—Newton's third law states that for every force there
is an equal and opposite reaction force
4.
What force is needed to keep a 60-N block moving across
asphalt ( = 0.67) in a straight line at a constant speed?
(A) 40 N
(B) 51 N
(C) 60 N
(D) 120 N
A—W/o acceleration then F = Ff.
Ff = Fn = mg Ff = (2/3)(60 N) = 40 N
5.
If F1 is the force exerted by the earth on the moon and F2 is
the force exerted by the moon on the earth, then which of
the following is true?
(A) F1 equals F2 and in the same direction
(B) F1 equals F2 and in the opposite direction
(C) F1 is greater than F2 and in the opposite direction
(D) F2 is greater than F1 and in the opposite direction
B—Newton's third law states that for every force there
is an equal, but opposite reaction force
6.
How far will a spring (k = 100 N/m) stretch when a 10-kg
mass hangs vertically from it?
(A) 1 m
(B) 1.5 m (C) 5 m
(D) 10 m
A—Fs = kx  mg = kx
(10 kg)(10 m/s2) = (100 N/m)x  x = 1 m
7.
Compared to the force needed to start sliding a crate
across a rough level floor, the force needed to keep it
sliding once it is moving is
(A) less
(B) greater
(C) the same
What is the frictional force, Ff, acting on the object?
(A) 5 N
(B) 10 N
(C) 20 N
(D) 40 N
B—Fnet = F – Ff = ma 
50 N – Ff = (4.0 kg)(10 m/s2)  Ff = 10 N
12. Two blocks are pushed along a horizontal frictionless
surface by a force, F = 20 N to the right.
F
The force that the 2-kg block exerts on the 3-kg block is
(A) 8 N to the left
(B) 8 N to the right
(C) 10 N to the left
(D) 12 N to the right
A—Force is proportional to the mass being pushed 
F2/F = 2/5  F2 = 2/5F = 2/5(20 N) = 8 N (left)
Questions 13-15 A box of mass m is on a ramp tilted at an
angle of  above the horizontal. The box is subject to the
following forces: friction (Ff), gravity (Fg), pull (Fp) and
normal (Fn). The box is moving up the ramp.
Fn
Fp
A—The coefficient k is less when the object is sliding
(k < s)  it takes less force to keep it sliding
8.
Which vector diagram best represents a cart slowing down
as it travels to the right on a horizontal surface?
(A)
(B)
(C)
(D)
Ff

Fg
13. Which formula is correct for Fn?
(A) Fn = Fg
(B) Fn = Fgcos
(C) Fn = Fgsin
(D) Fn = Fgcos
D—Fn equals the component of gravity that is
perpendicular to the surface: Fgcos
B—Slowing down the force of friction, Ff, must be
greater than the push/pull force, F
9.
A skier on waxed skis (k = 0.05) is pulled at constant
speed across level snow by a horizontal force of 40 N.
What is the normal force exerted on the skier?
(A) 700 N (B) 750 N (C) 800 N (D) 850 N
No acceleration then F = Ff = Fn 
40 N = (0.05)Fn  Fn = 800 N
14. Which formula is correct for Ff if the box is accelerating?
(A) Ff = Fp
(B) Ff = Fp – Fgsin
(C) Ff = Fgcos
(D) Ff = Fn
D—Ff = Fn
15. Which formula is correct for Fp if the box is moving at
constant velocity?
(A) Fp = Ff
(B) Fp = Ff + Fgsin
(C) Fp = ma - Ff
(D) Fp = Ff – Fgsin
B—Fnet = ma = 0
Fp – Ff – Fgsin = 0  Fp = Ff + Fgsin
16. A rope supports a 3-kg block. The breaking strength of the
rope is 50 N. The largest upward acceleration that can be
given to the block without breaking the rope is most nearly
(A) 6 m/s2 (B) 6.7 m/s2 (C) 10 m/s2 (D) 15 m/s2
B—Fnet = Fp – Fg = ma
50 N – (3 kg)(10 m/s2) = (3 kg)a  a = 20/3 = 6.7 m/s2
17. A block of mass 3m can move without friction on a
horizontal table. This block is attached to another block of
mass m by a cord that passes over a frictionless pulley.
The masses of the cord and the pulley are negligible,
A—Ff = Fn
Ff = (mg + Tsin)
Questions 22-23 A block of mass m is accelerated across a
rough surface by a force F that is exerted at an angle
with the horizontal. The frictional force is f.
22. What is the acceleration of the block?
(A) F/m
(B) Fcos/m
(C) (F – f)/m
(D) (Fcos – f)/m
D—Fnet = Fcos – f = ma
a = (Fcos – f)/m
What is the acceleration of the descending block?
(A) Zero
(B) ¼ g
(C) ⅓ g
(D) ⅔ g
B—Fnet = ma (Fnet is Fg on m, and m = total mass = 4 m)
mg = (4m)a  a = ¼ g
18. A 5-kg block lies on a 5-m inclined plane that is 3 m high
and 4 m wide. The coefficient of friction between the plane
and the block is 0.3.
What minimum force F will pull the block up the incline?
(A) 12 N
(B) 32 N
(C) 42 N
(D) 50 N
C—Fnet = Ff + Fg-||
Fnet = 4/5mg) + 3/5mg = (0.3)(40 N) + 30 N = 42 N
19. A 10-kg block is pulled along a horizontal surface at
constant speed by a 50-N force, which acts at a 37o angle
with the horizontal.
What is the coefficient of friction ?
(A) 1/3
(B) 2/5
(C) 1/2
(D) 4/7
D—Fnet = Ff = Fn
(50 N)cos37 = [(10 kg)(10 m/s2) – (50 N)sin37]  = 4/7
20. A 10-kg cart moves without frictional loss on a level table.
A 50-N force pulls horizontally to the right (F1) and a 40-N
force at an angle of 60° pulls on the cart to the left (F2).
F2
23. What is the coefficient of friction between the block and the
surface?
(A) f/mg
(B) mg/f
(C) (mg – Fcos)/f
(D) f/(mg – Fsin)
D—f = Fn = (mg – Fsin)
 = f/(mg – Fsin)
24. A horizontal force F pushes a block of mass m against a
vertical wall. The coefficient of friction between the block
and the wall is .
What minimum force, F, will keep the block from slipping?
(A) F = mg
(B) F = mg
(C) F = mg/
(D) F = mg(1 – )
C—Fnet = Ff – Fg = 0
0 = Fn – mg = F – mg  F = mg/
25. Two blocks of masses M and m, with M > m, are connected
by a light string. The string passes over a frictionless
pulley of negligible mass so that the blocks hang vertically.
The blocks are then released from rest. What is the
acceleration of the block of mass M?
(A) g
(B) (M – m)g/M
(C) (M + m)/Mg
(D) (M – m)g/(M + m)
D—Fnet = ma
(M – m)g = (M + m)a  a = (M – m)g/(M + m)
26. The momentum vector for an object is represented below.
The object strikes a second object that is initially at rest.
Which set of vectors may represent the momentum of the
two objects after the collision?
(A)
(B)
F1
What is the horizontal acceleration of the cart?
(A) 1 m/s2 (B) 2 m/s2 (C) 3 m/s2 (D) 4 m/s2
C—Fnet = F1 – F2cos60 = ma (cos60 = ½)
50 N – 40 N(½) = (10 kg)a  a = 3 m/s2 (to the right)
21. A push broom of mass m is pushed across a rough floor by
a force T directed at angle . The coefficient of friction is .
The frictional force on the broom has magnitude
(A) (mg + Tsin)
(B) (mg - Tsin)
(C) (mg + Tcos)
(D) (mg - Tcos)
(C)
(D)
C—When you place the two vectors tail to tip, only c
will equal the original vector.
Questions 27-28 A 100-N weight is suspended by two cords.
27. The tension in the ceiling cord is
(A) 50 N
(B) 100 N (C) 170 N
(D) 200 N
D—TL, Fg and TR make a 30-60-90 triangle, where
sin30 = Fg/TR  TR = (100 N)/sin30 = 100 N/½ = 200 N
28. The tension in the wall cord is
(A) 50 N
(B) 100 N (C) 170 N
3.
(D) 200 N
C—tan30 = Fg/TL
 TL = (100 N)/tan30 = 100 N/(1/3) = 170 N
1.
d = vot + ½at2
55 m = (25 m/s)(3.0 s) + ½a(3.0 s)2 a = -4.4 m/s2
Practice Free Response
Block A (8 kg) is pulled along a frictionless table by block B
(2 kg) hanging over the table's edge.
a. Label all the forces acting on the two blocks.
Fn
Ff
A 4700 kg truck carrying a 900 kg crate is traveling at 25
m/s to the right along a straight, level highway. The driver
then applies the brakes, and as it slows down, the truck
travels 55 m in the next 3.0 s. The crate does not slide.
a. Calculate the magnitude of the acceleration of the
truck, assuming it is constant.
b.
Label all the forces acting on the crate during braking.
Fn
Ff
Ft-A
c.
Ft-B
Fg-A
Fg
Calculate the minimum coefficient of friction between
the crate and truck that prevents the crate from sliding.
F = Ff = Fn
ma = mg   = a/g = 4.4 m/s2/10 m/s2 = 0.44
b.
Now assume the bed of the truck is frictionless, but there is
a spring of spring constant 9200 N/m attaching the crate to
the truck. The truck is initially at rest.
Fg-B
Determine the acceleration of the system.
Fnet = ma
20 N = (8 kg + 2 kg)a  a = 2 m/s2
c.
Determine the tension on the 8-kg block.
Ft = ma = (8 kg)(2 m/s2) = 16 N
d.
d.
Determine the tension on the 2-kg mass.
Fg – Ft = ma
(2 kg)(10 m/s2) – Ft = (2 kg)(2 m/s2)  Ft = 16 N
2.
An empty sled of mass 25 kg slides down a snowy hill with
a constant speed of 2.4 m/s. The slope of the hill is
inclined at an angle of 15° with the horizontal.
a. Calculate the time it takes the sled to go 21 m down
the slope.
d = vt
21 m = (2.4 m/s)t  t = 8.75 s
b.
c.
4.
Label all the forces acting on the sled.
Fn
Ff
Fg
Calculate the force of friction.
Fg = Ff  (mbucket + msand)g = smblockg
1.35 kg + msand = (0.450)(28.0 kg)  msand = 11.25 kg
Fnet = Fgsin – Ff = 0
Ff = (25 kg)(10 m/s2)sin15o = 65 N
d.
Calculate the coefficient of friction.
Ff = Fn = (Fgcos)
65 N = (25 kg)(10 m/s2)cos15o  = 0.27
e.
If the truck and crate have the same acceleration,
calculate the extension of the spring as the truck
accelerates from rest to 25 m/s in 10 s.
a = (v – vo)/t = 25 m/s/10 s = 2.5 m/s2
F = Fs = kx  ma = kx
x = ma/k = (900 kg)(2.5 m/s2)/(9200 N/m) = 0.24 m
e. What happens to the spring when the truck reaches 25
m/s and stops accelerating? Explain your reasoning.
The spring shortens. Without acceleration, the spring
force is not needed to keep the crate on the truck bed
so the spring will go to its unstretched length.
A 28.0-kg block rests on a table
(s = 0.450, k = 0.320). The block
is connected to an empty 1.35-kg
bucket by a cord running over a
frictionless pulley. Sand is gradually
added to the bucket until the system
just begins to move.
a. Calculate the mass of sand added to the bucket.
5.
The sled reaches the bottom of the hill and continues
on horizontal ground. Graph velocity vs. time for the
sled until it stops (Assume the same ). Indicate the
time t when the sled reaches the horizontal ground.
velocity
b. Calculate the acceleration of the system.
Fnet = Fgbucket + Fgsand – Ffblock = mtotala
(mbucket + msand – kmblock)g = (mbucket + msand + mblock)a
a = [12.6 kg – (0.320)(28.0 kg)]10 m/s2]/40.6 kg = 0.90 m/s2
A 7650-kg helicopter accelerates upward at 0.80 m/s 2
while lifting a 1250-kg frame at a construction site.
a. What is the lift force of the helicopter rotors?
Fp = mg + ma = m(g + a)
Fp = (7650 kg + 1250 kg)(10.80 m/s2) = 96,100 N
b.
What is the tension in the cable (ignore its mass) that
connects the frame to the helicopter?
Fnet = ma  Ft – mg = ma
Ft = m(a + g) = (1250 kg)(0.80 m/s2 + 10 m/s2) = 13,500 N
t
time
c. What force does the cable exert on the helicopter?
Fp – Fg –Ft = ma
96,100 N – 76500 N – Ft = (7650 kg)(0.80 m/s2)
Ft = 13,500 N
6.
A rope of negligible mass passes over a pulley of
negligible mass attached to the ceiling. One end of the
rope is held by Student A of mass 70 kg, who is at rest on
the floor. The opposite end of the rope is held by Student B
of mass 60 kg, who is suspended at rest above the floor.
a. Draw and label all the forces on Student A and on
Student B.
TB
TA
Fg-B
b.
Fg-A
Calculate the magnitude of the force exerted by the
floor on Student A.
Fn = Fg-A – Fg-B = mAg – mBg = (mA – mB)g
Fn = (70 – 60)(10) = 100 N
Student B now climbs up the rope at a constant
acceleration of 0.25 m/s2 with respect to the floor.
c. Calculate the tension in the rope while Student B is
accelerating.
Fnet = TB – Fg-B = ma
TB = mBg + mBa = mB(g + a) = (60)(10 + 0.25) = 615 N
d.
As Student B is accelerating, is Student A pulled
upward off the floor? Justify your answer.
No, the upward force TA = 615 N < the downward force
Fg-A = 700 N  the net force on student A is downward.
e.
With what minimum acceleration must Student B climb
up the rope to lift Student A upward off the floor?
Fnet = TB – FgB = mBa (TB = TA = mAg)
mAg – mBg = mBa
a = (mA – mB)g/mB = (70 – 60)10/60 = 1.7 m/s2
f. How could student B's acceleration up the rope
(measured from the floor) be increased?
Anchor student A with additional weight.
7.
A 100-kg box is anchored to the ceiling by two cords.
50o
25o
Determine the tension in each cord.
TLcos130 + TRcos25 = 0
TL(-0.643) + TR(0.906) = 0  TR = TL(0.710)
TLsin130 + TRsin25 + mgsin-90 = 0
TL(0.766) + TL(0.710)(0.423) + (100 kg)(10 m/s2)(-1) = 0
TL(1.07) = 1000 N  TL = 938 N
TR = TL(0.710) = (938 N)(0.710) = 666 N