Force and Newton’s Laws

The study of motion is called kinematics. The study of the causes of motion (and changes in motion) is called dynamics.

In the study of dynamics, there are three essential quantities to understand:

1) mass measure of amount of matter

2) force a push or pull that may change the motion of an object3) acceleration the timed rate change of the motion of an object.

Newton’s Laws of Motion

Note: We will still consider a moving object as a single particle!

Newton’s Law of Inertia: An object’s motion will not change unless a net force acts upon it.

• this will be true for inertial frames of reference-- frames that are not considered to be accelerating in relation to one another.

Newton’s Law of Acceleration: Net forces produce acceleration that is directly related to the force and inversely related to the mass of the object.

∑F = ma , so ∑Fx = max , ∑Fy = may , ∑Fz = maz

A student pushes a 240 kg sled for a distance of 2.3 m over the surface of a frictionless (how the hell does he push it?), frozen lake. What velocity will the sled have if she uses a constant force of 130 N while pushing from rest?

m = 240 kg ∆x = 2.3 m

Fx = 130 Nvo = 0

v = ?

v = √ vo2 + 2a∆x a = F / m

= .54 m/s2

= 1.6 m/s

What constant force does the student need to use in order to reverse the direction of the velocity of the sled in 4.5s?

F = ma a = v - vo

t

= - 0.71 m/s2

= - 170 N

v0 = 1.6 m/s

v = - 1.6 m/s

Newton’s Law of Interaction: To every action there is an equal an opposite reaction.

• a “single” force does not exist

• a single force isolated in problems is always part of a pair of forces acting between two bodies

Weight: a very specific force caused by the effect of gravity on a mass. W = mg

“Weightless” is only a perception-- it can only really be achieved in deep space where g is negligible.

Normal and Lift ForcesNormal force acts perpendicular to the surface:

Opposite to mg on a level surface, or

opposite mgcosø on an incline:

Normal and Weight are not Newton Action-Reaction pairs– Why not?

Lift acts in the same manner for an aircraft in flight

30˚ 45˚

15.0 kg

Find the tension in each string.

T1T2

T3T3

mg∑Fy= 0 = T3 - mg

T3 = mg

∑Fx = max = 0 = T2x - T1x

∑Fy = may = 0 = T1y + T2y - T3

cos(45)T2 - cos(30)T1 = .707T2 - .866 T1 = 0

sin(30)T1 + sin(45)T2 - mg = 0

.500T1 + .707T2 = 147

T2 = 1.22T1

.500T1 + .707(1.22)T1 = 1471.36T1 = 147

T1 = 108 N

T2 = 1.22(108) = 132N

A sled of mass 7.5 kg is pulled along a frictionless surface by a rope that is angled 15˚ above the horizontal. What is the acceleration of the sled if the tension in the rope is 21.0 N?

mg

N T

ø

∑Fx = max = Tcosø

∑Fy = may = 0 = Tsinø + N - mg

a = Tcosø

m= (21.0 N)(cos 15˚)

7.5 kg

= 2.70 m/s2

An 18.0 kg block is held at the top of a 27˚ frictionless incline by a rope. Find the tension in the rope and what is the speed of the block at the bottom of the 4.00 m ramp if the rope breaks?

mg

T

ø

N

ømgcosø

mgsinø

∑Fx = max = 0 = T - mgsinø

∑Fy = may = 0 = N - mgcosø

T = mgsinø = (18.0)(9.8)(sin27˚)

= 80.0 N

If rope breaks: ∑Fx = max = 0 - mgsinø

a = - mgsinø/m = - gsinø = - 4.44 m/s2

v = √ vo 2 + 2a∆x

Note that the acceleration is negative -- the object is traveling in the negative x direction!

∆x = - 4.00 m

= - 5.96 m/s

The answer to this equation can be +/- so choose the - value because it travels in the - x direction!

v0 = 0

Elevator Problems

N

mg

If the elevator has an upward a, then:

∑Fy = ma = N - mg

N = m(a + g)

If the elevator has a downward a, then:

∑Fy = ma = mg - N

N = m(g - a)In freefall:a = g

therefore, N = 0

Connected Masses / Atwood’s Machine

1

2

Block 1 of mass 35.0 kg sits on a frictionless surface connected to block 2 of mass 15.0 kg by a cord passed over a frictionless, massless pulley:

Find A) the acceleration of the blocks and B) the tension in the cord.

m1g

N

T1 m2g

T2

For Block 1:

∑Fx = m1a1x = T1

∑Fy = 0 = N - m1g

For Block 2:

∑Fy = m2a2y = m2g - T2

Because they are connected and the pulley is ideal: a1x = a2y = a …and: T1 = T2 = T

m2a = m2g - m1a

a = m2g

(m1 + m2)

= 2.94 m/s2

T = m1a

= 103 N

1

2

Atwood’s Machine

m1 = 12.0 kg

m2 = 21.0 kg

Find A) the acceleration of the each of the blocks and B) the tension in the rope. (Pulley is ideal!)

m1g

T1

m2g

T2

∑Fy1 = T1 - m1g = m1a1

∑Fy2 = m2g - T2 = m2a2

Since the blocks are connected by an ideal pulley

T1 = T2 = T and a1 = a2 = a

T = m1a1 + m1g m2a = m2g - (m1a + m1g)

a = (m2 - m1)g

m2 + m1

= 2.67 m/s2

1: 2.67 m/s2

2: -2.67 m/s2