ObservationsThe acceleration of the given body is directly

proportional to the applied force. This means that the ratio of the acceleration is always constant.

— = — = — = constantF1 F2 F3

a1 a2 a3The constant is a measure of how effective is the given force in producing acceleration. This ratio is the property of the body called mass.

m = — Fa

A force of 1 newton (1N) is that resultant force that will give a 1-kg mass an acceleration of 1 m/s2. The newton (N) is the SI unit of force.

Newton’s 2nd Law of MotionApplying a constant force of 12 N in succession to 1-, 2- and 3-kg masses will produce accelerations of 12 m/s2, 6 m/s2 and 4m/s2, respectively.

a = —— = —— = 12 m/s2 F 12Nm1 1kg

a = —— = —— = 6 m/s2

a = —— = —— = 4 m/s2

F 12Nm2 2 kgF 12Nm3 3kg

Newton’s Second Law of Motion: “Whenever an unbalanced force acts on a body, it produces in the direction of the force an acceleration that is directly proportional to the force and inversely proportional to the mass of the body.”

ΣF = + F1 + F2 + F3 + …

Force (N) = mass (kg) x acceleration (m/s2)Force (lb) = mass (slug) x acceleration (ft/s2)

1 lb = 4.448 N 1 slug = 14.59 kg

Fnet = ma

Illustrative Example 2It is determined that a resultant force of 60

N will give a wagon an acceleration of 10 m/s2. What force is required to give the wagon an acceleration of 2 m/s2?

F = ma = (6kg)(2 m/s2)

m = —— = ——— = 6 kgF 60Na 10m/s2

F = 12 N

Illustrative Example 3A 1000kg car moving north at 100 km/h

brakes to a stop in 50 m. What are the magnitude and direction of the force?

Given: m = 1000kg; vi = 100 km/h = 27.8 m/s;

vf = 0 Find: Fnet = ? Formula: Fnet = ma

a = ———vf – vi

ΔtΔX 50 m

v 13.9 m/sΔt = —— = ————

Δt = 3.6 sa = ————0 – 27.8 m/s

3.6 s

a = – 7.72 m/s2

F = ma = (1000kg)(- 7.72 m/s2)

F = 7720 N, South

Therefore, we can summarize as follows:

SI: W(N) = mkg) x g(9.8m/s2)

English: W(lb) = m(slug) x g(32 ft/s2)

Relationship Between Mass and WeightMASS is a universal constant equal to the ratio of the body’s

weight to the gravitational acceleration due to gravity.WEIGHT is the force of gravitational attraction and varies

depending of the acceleration due to gravity.

W = mg or m = ——Wg

(1)The mass of a particle is equal to its weight divided by the acceleration due to gravity.

(2)Weight has the same units as the unit of force.

(3)The acceleration of gravity has the same units as acceleration.

Illustrative Examples

What is the weight of a 4.8-kg mailbox?

What is the mass of a 40-N tank?

What is the mass of a 60-lb child?

What is the weight of a 7-slug man? W = mg = (4.8kg)(9.8 m/s2)= 47Nm = F/g = (40N)/(9.8 m/s2)= 4.08 kg

m = F/g = (60lb)/(32 ft/s2)= 1.9 slugW = mg = (7slug)(32 m/s2)= 224 lb

Illustrative Example 2

Find the weight of the body whose weight on Earth is 100 N. If this mass were taken to a distant planet where g = 2.0 m/s2, what would be its weight on that planet?

Given: WE = 100N; gE = 9.80 m/s2

gP = 2.0 m/s2

Find: WP = ?

Solution

Mass on Earth:

m = —— = ———— = 10.2 kg WE 100 N

gE 9.80 m/s2

WP = mgP = (10.2 kg)(2 m/s2)

Weight on the planet:

WP = 20.4 N Ans.: a = 1.63 m/s2; m = 81.6 kg on both places

Try this!A woman weighs 800 N on

Earth. When she walks on the moon, she weighs only 133 N. What is the acceleration due to gravity on the Moon, and what is her mass on the Moon? On the Earth?

1. A ball is accelerated from rest at a rate of 1.20 m/s2 after a force of 20.0 n is applied. What is the mass of the ball?

Practice Exercise (p.61-62)

2. A 15.0-kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box?

3. If the forces in number 2 are applied opposite each other, what is the magnitude and direction of the acceleration?

Solution

Solution

Solution

Solution:1. A ball is accelerated from rest at a rate of 1.20 m/s2 after a

force of 20.0 n is applied. What is the mass of the ball?

Given: Fnet = 20.0 N; Find: m = ?

a = 1.20m/s2

m = —— = ————— = 16.7 kg Fa

20.0 N1.20 m/s2

F = ma

Solution:

Given: m = 15.0 kg; F1 = +15.0 N; F2 = +18.0 N

Find: a = ?

a = —— = ———————— Fnet

m15.0 N + 18.0 N

15.0 kg

Fnet = F1 + F2 = ma

2. A 15.0kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box?

a = 2.20 m/s2

Solution:

Given: m = 15.0 kg; F1 = +15.0 N; F2 = +18.0 N

Find: a = ?

a = —— = ———————— Fnet

m15.0 N + 18.0 N

15.0 kg

Fnet = F1 + F2 = ma

2. A 15.0kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box?

a = 1.53 m/s2 to the right

Solution:

Given: m = 15.0 kg; F1 = +15.0 N; F2 = -18.0 N

Find: a = ?

a = —— = ———————— Fnet

m15.0 N - 18.0 N

15.0 kg

Fnet = F1 + F2 = ma

3. If the forces in number 2 are applied opposite each other, what is the magnitude and direction of the acceleration?

a = 0.2 m/s2 to the left

mM

The Atwood Machine

T TM m

T T

Mg mg

a a

The Atwood Machine

M m

T T

Mg mg

a a

For mass M:

T – Mg = - MaMg - T = Ma Eq. 1

For mass m:

T – mg = ma Eq. 2

Conbining eq.1 & eq. 2:

Mg - T = MaT – mg = ma

+

Mg - mg = (M + m)a

a = ——— g M – m

M + m

Example: Atwood Machine

• A 2.0-kg body and a 5.0-kg body are each suspended at the end of a cord that passes over a frictionless pulley. (a) What is the acceleration of the system? (b) What is the tension on the cord?

5.0 kg2.0 kg aa

TT

Vertical and horizontal problems

m1

m2

A 2.00-kg hanging block pulls a 3.00-kg block along a frictionless table. Calculate for the acceleration of the system and the tension on the cord.

Vertical and horizontal problems

m1

m2

m1

m2

T

a

w2

aT

Fnet = T = m1a - eq.1

Fnet = T- w2 = -m2a or w2 – T = m2a - eq.2

T

T

Since the blocks are connected by a single cord, the tension T of the cord for both blocks is the same, thus similar rate of motion for both.

T = m1a w2 – T = m2a

Combining the 2 equations gives:

w2 = m1a + m2aw2 = (m1 + m2)a

a = ———————w2

(m1 + m2)Working equation

---- Eq. 1

---- Eq. 2

a = ———————w2

(m1 + m2)

a = ———————(2.00kg)(9.80 m/s2)

(3.0 kg + 2.0 kg)

a = 3.92 m/s2

For acceleration:

For the tension: from eq.1

T = m1a = (3.0 kg)(3.92m/s2)= 11.8 N

Another Example

Example: Horizontal and Vertical motion

• A 100-g mass lies on a frictionless table and a cord is attached to it. The cord passes over a pulley at the edge of the table and at the free end, a 10-g mass is hung. Find the acceleration of the system and the tension on the cord.

100 g

10 g

Solution

a = ——————w2

(m1 + m2)

a = ———————— 10g (980cm/s2)

(100g + 10g)

a = 89.1 cm/s2

The acceleration: The tension:

Isolating the 10-g mass, the tension T acts upward and the acceleration downward. The unbalanced force F = (9800 – T)dynes. Applying the second law:

F = ma 9800-T = 10g(89.1cm/s2)

T = 9800dy – 891dy

T = 8910 dynes

• A 100-g mass lies on a frictionless table and a cord is attached to it. The cord passes over a pulley at the edge of the table and at the free end, a 10-g mass is hung. Find the acceleration of the system and the tension on the cord.

m1

m2 m3

T1T2

T2 T1

Vertical and horizontal problems

If a 100-g counter weight is attached on the left side of m1 which is 300 g, and a 200-g mass on the right, what would be the acceleration of the system and the tensions on the left and right cords?

m1 T1T2

Free-body Diagrams

m2 m3

T2 T1

w3 w3

Fnet = T1 – T2 = m1a

Fnet = T2 – W2 = m2aFnet = T1 – W3 = -m3a

aa

m1 = 300 gm2 = 100gm3 = 200 g

Combining the 3 equations yields:

T1 – T2 = m1a

T2 – W2 = m2a

W3 – T1 = m3a

W3 – W2 = (m1 + m2 + m3)a

a = ——————W3 – W2

m1 + m2 + m3

+

+

Working equation

Fnet2 = T2 – W2 = m2a

Fnet1 = T1 – T2 = m1a

Fnet3 = T1 – W3 = -m3a

a = —————

= —————————————

W3 – W2

m1 + m2 + m3

a = 1.63 m/s2

(0.2kg)(9.8m/s2) – (0.1kg)(9.8m/s2)

0.3 kg + 0.2 kg + 0.1 kg

m1 = 300 gm2 = 100gm3 = 200 g

m1 = 300 gm2 = 100gm3 = 200 g

T1 – T2 = m1a –Eq.1

T2 – W2 = m2a –Eq.2

W3 – T1 = m3a –Eq.3For Tensions, T1 and T2For Tensions, T1 and T2

W3 – T1 = m3am3g – T1 = m3aT1 = m3g – m3aT1 = m3(g – a)T1 = 0.2kg(9.8m/s2 – 1.63m/s2)T1 = 1.63 N

Using Equation 3

Using Equation 2T2 – W2 = m2aT2 – m2g = m2aT2 = m2a + m2g T2 = m2(g – a)

T2 = 0.1kg(9.8m/s2 + 1.63m/s2)

T2 = 1.14 N

Using Equation 1

T1 – T2 = m1a

1.63N – T2 = 0.3kg(1.63m/s2)

1.63N – T2 = 0.489 N

T2 = 1.63N – 0.489 N

T2 = 1.14 N

T1 = 1.63 N

Solution

a = ——— g M – m

M + m

a = ————— (9.8m/s2) 5kg – 2kg

5kg + 2kg

a = 4.2 m/s2

Another Solution

5.0 kg2.0 kg aa

F = maUnbalanced force F:

F = W2 + W1 = 5kg(9.8m/s2) – 2kg(9.8m/s2)

= 49 N – 19.6 N = 29.4N

The moving masses:

m = 2kg + 5kg = 7kg

T T

The acceleration: F = ma 29.4N = 7kg (a) a = 29.4 N / 7kg = 4.2 m/s2

w1 w2

Solution: The tension on the cord

Consider the 2.0-kg body as the moving part of the system. We can isolate it as shown in the free-body diagram on the right. Let T be the tension on the cord. The unbalanced force is T – W1. The acceleration is 4.2

m/s2.

F = ma T – 19.6N = (2.0kg)(4.2m/s2)

2.0 kg

T

a

T = 8.4 N + 19.6 N = 28.0 N