Chapter 6

Force = mass X acceleration
S
F = ma
Sum of all the forces acting on a body
Vector quantity
S
F = 0
-Still object or
-Obj. at constant velocity
S
F = ma
-Accelerating object

Situation One:
Non-moving Object
• Still has forces
Force of the material of the rock
Force of gravity http://alfa.ist.utl.pt/~vguerra/Other/Rodin/thinker.jpg

Situation Two: Moving Object: Constant Velocity
S
F = 0
F pedalling
= F air
+ F friction
F friction
F air
F pedalling http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg

Situation Two: Moving Object: Accelerating
S
F = ma ma = F pedalling
– F air
- F friction
F friction
F air
F pedalling http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg

Unit of Force = the Newton
S
F=ma
S
F = (kg)(m/s 2 )
1 N = 1 kg-m/s 2  (MKS)
1 Newton can accelerate a 1 kg object from rest to 1 m/s in 1 s.

1. No motion
2. Constant velocity
BOTH INDICATE NO ACCELERATION
S
F=0

Three ropes are tied together for a wacky tug-ofwar. One person pulls west with 100 N of force, another south with 200 N of force.
Calculate the magnitude and direction of the third force.
100 N
200 N
?

A car with weight 15,000 N is being towed up a
20 o slope (smooth) at a constant velocity. The tow rope is rated at 6000 N. Will it break?

• S
F=ma
• Must add up all forces on the object

What Force is needed to accelerate a 5 kg bowling ball from 0 to 20 m/s over a time period of 2 seconds?

Calculate the net force required to stop a 1500 kg car from a speed of 100 km/h within a distance of 55 m.
100 km/h = 28 m/s v 2 = v o
2 + 2a(x-x o
) a = (v 2 - v o
2 )/2(x-x o
) a = 0 2 – (28 m/s) 2 /2(55m) = -7.1 m/s 2

A 1500 kg car is pulled by a tow truck. The tension in the rope is 2500 N and the 200 N frictional force opposes the motion. The car starts from rest.
a. Calculate the net force on the car b. Calculate the car’s speed after 5.0 s

A 500.0 gram model rocket (weight = 4.90 N) is launched straight up from rest by an engine that burns for 5 seconds at 20.0 N.
a. Calculate the net force on the rocket b. Calculate the acceleration of the rocket c. Calculate the height and velocity of the rocket after 5 s d. Calculate the maximum height of the rocket even after the engine has burned out.

Calculate the sum of the two forces acting on the boat shown below. (53.3 N, +11.0
o )

Mass
– The amount of matter in an object/INTRINSIC
PROPERTY
– Independent of gravity
– Measured in kilograms
Weight
– Force that results from gravity pulling on an object
– Weight = mg (g = 9.8 m/s 2 )

• Weight = mg is really a re-write of F=ma.
– Weight is a force
– g is the acceleration (a) of gravity
• Metric unit of weight is a Newton
• English unit is a pound

A 60.0 kg person weighs 1554 N on Jupiter. What is the acceleration of gravity on Jupiter?

a= 0
S
F = F
N
– mg ma = F
N
– mg
0 = F
N
– mg
F
N
= mg
Suppose Chewbacca has a mass of 102 kg:
F
N
= mg = (102kg)(9.8m/s 2 )
F
N
= 1000 N
F
N mg a is zero

a = 4.9 m/s 2
S
F = F
N
– mg ma = F
N
– mg
F
N
= ma + mg
F
N
= m(a + g)
F
N
=(102kg)(4.9m/s 2 +9.8 m/s 2 )
F
N
= 1500 N
F
N mg a is upward

a = -4.9 m/s 2
S
F = F
N
– mg ma = F
N
– mg
F
N
= ma + mg
F
N
= m(a + g)
F
N
=(102kg)(-4.9m/s 2 +9.8 m/s 2 )
F
N
= 500 N
F
N mg a is down

At what acceleration will he feel weightless?
F
N
= 0
S
F = F
N
– mg ma = F
N
– mg ma = 0 – mg ma = -mg a = -9.8 m/s 2
Apparent weightlessness occurs if a > g
F
N mg

A 10.o kg present is sitting on a table. Calculate the weight and the normal force.
F
N
F g
= W

Suppose someone leans on the box, adding an additional 40.0 N of force.
Calculate the normal force.

Now your friend lifts up with a string (but does not lift the box off the table). Calculate the normal force.

What happen when the person pulls upward with a force of 100 N?
S
F = F
N
+ F p
– mg
S
F = 0 +100.0N – 98N = 2.0N
F p
= 100.0 N ma = 2N a = 2N/10.0 kg = 0.2 m/s 2
F g
= mg = 98.0 N

A person pulls on the box (10.0 kg) at an angle as shown below. Calculate the acceleration of the box and the normal force. (78.0 N)
F p
= 40.0 N
30 o
F
N mg

F fr
• Always opposes the direction of motion.
• Proportional to the Normal Force (more massive objects have more friction)
F
N
F a
F
N mg mg

Static -opposes motion before it moves
( m s
)
• Generally greater than kinetic friction
• F max
= Force needed to get an objct moving
F max
= m s
F
N
Kinetic - opposes motion while it moves
( m k
)
• Generally less than static friction
F fr
= m k
F
N

Rolling uses static friction
– A new part of the wheel/tire is coming in contact with the road every instant
A
B

Braking uses kinetic friction
A
Point A gets drug across the surface

A 50.0 kg wooden box is pushed across a wooden floor ( m k
=0.20) at a steady speed of
2.0 m/s.
a. How much force does she exert? (98 N) b. If she stops pushing, calculate the acceleration. (-1.96 m/s 2 ) c. Calculate how far the box slides until it stops.
(1.00 m)

A 100 kg box is on the back of a truck ( m s
The box is 50 cm X 50 cm X 50 cm.
= 0.40). a. Calculate the maximum acceleration of the truck before the box starts to slip.

Your little sister wants a ride on her sled. Should you push or pull her?

What trigonometric function does this resemble?

q
F
N mg

q
F
N mg q mgcos q mgsin q

q
F
N
F fr mgsin q mgcos q

A 50.0 kg file cabinet is in the back of a dump truck ( m s
= 0.800).
a. Calculate the magnitude of the static friction on the cabinet when the bed of the truck is tilted at 20.0
o (170 N) b. Calculate the angle at which the cabinet will start to slide. (39 o )

Given the following drawing: a. Calculate the acceleration of the skier. (snow has a m k of 0.10) (4.0 m/s 2 ) b. Calculate her speed after 4.0 s? (16 m/s)

First we need to resolve the force of gravity into x and y components:

F
Gy
= mgcos30 o
F
Gx
= mgsin30 o
The pull down the hill is:
F
Gx
= mgsin30 o
The pull up the hill is:
F fr
= m k
F
N
F fr
= (0.10)(mgcos30 o )

S
F = pull down – pull up
S
F = mgsin30 o – (0.10)(mgcos30 o ) ma = mgsin30 o – (0.10)(mgcos30 o ) ma = mgsin30 o – (0.10)(mgcos30 o ) a = gsin30 o – (0.10)(gcos30 o ) a = 4.0 m/s 2
(note that this is independent of the skier’s mass)

To find the speed after 4 seconds: v = v o
+ at v = 0 + (4.0 m/s 2 )(4.0 s) = 16 m/s

Suppose the snow is slushy and the skier moves at a constant speed. Calculate m k
S
F = pull down – pull up ma = mgsin30 o – ( m k
)(mgcos30 o ) ma = mgsin30 o – ( m k
)(mgcos30 o ) a = gsin30 o – ( m k
)(gcos30 o )

Since the speed is constant, acceleration =0
0 = gsin30 o – ( m k
)(gcos30 o )
( m k
)(gcos30 o ) = gsin30 o m k
= gsin30 o = sin30 o = tan30 o =0.577
gcos30 o cos30 o

D ≈ ¼Av 2
D = drag force
A = Area
V = velocity
Fails for
• Very small particles (dust)
• Very fast (airplanes)
• Water and dense fluids

Derive the formula for the acceleration of a freefalling object including the drag force.

a. Find the formula for terminal speed (a=0) for a freefalling body b. Calculate the terminal velocity of a person who is 1.8 m tall, 0.40 m wide, and 75 kg.
(64 m/s)

A 1500 kg car is travelling at 30 m/s when the driver slams on the brakes ( m k
= 0.800).
Calculate the stopping distance: a. On a level road. (57.0 m) b. Up a 10.0
o incline (48.0 m) c. Down a 10.0
o incline (75.0 m)

A dogsled has a mass of 200 kg. The sled reaches cruising speed, 5.0 m/s in 15 m. Two ropes are attached to the sled at 10.0o, one on each side connected to the dogs. ( m k
= 0.060) a. Calculate the acceleration of the sled. (0.833 m/s) b. Calculate T
1 and T period. (140 N)
2 during the acceleration

S
F=ma
Weight = mg (g = 9.8 m/s 2 )
F max
F fr
= m s
F
N
= m k
F
N