Further applications of Newton’s laws

m1

m2

massless pulley, rope

For the circumstance to the right suppose m2 > m1.

We chose this “direction” of acceleration as being positive.

When the masses are released m2 will accelerate downwards, while m1 will accelerate upwards.

a

a

This means that acceleration up on the left is positive, but acceleration down on the right is positive.

This is required by the constraint of the rope that dictates that the when one accelerates up the other accelerates down.

Atwood’s Machine

m1

m2

T

– T

m2g

– m1g

Having chosen the signs of the of acceleration on the two sides also fixes the signs of the forces on the two sides.

On the left we have the tension, T, from the rope acting upward as positive and the force of gravity acting downward coming in as negative: –m1g (as usual).

On the right, however, since acceleration downward is positive, forces downward are positive while those upward are negative. So there we have –T for the upward tension in the rope and +m2g for the force of gravity.

a

a

2 11 1

2 1

(m m )T m g m g(m m )

−− =

+

And using this in the expression for m1 allows solving for T,

1 2

2 1

2m mT g(m m )

=+

Adding the equations eliminates T

2 1 1 2m g m g m a m a− = +

2 1 1 2(m m )g (m m )a− = +

2 1

2 1

(m m )a g(m m )

−=

+

For m1: 1 1 1F T m g m a= − =∑ For, m2: 2 2 2F T m g m a= − + =∑

With these definitions Newton’s 2nd law becomes:

m1

m2

T

– T

m2g

– m1g

Now solve for a,

2 1

2 1

(m m )a g(m m )

−=

+1 2

2 1

2m mT g(m m )

=+

m1

m2

T

– T

m2g

– m1g

,

Do these make sense?

i.e. m2 accelerates down with a = g.

If instead m2 = 0, a = – g , i.e. m1 accelerates down at the right rate & T = 0. Good.

If m1= m2= m, a = 0, and T = mg. Also good.

If m1 = 0, the first eqn. gives, 2

2

(m 0)a g g(m 0)

−= =

+

The second eqn gives, for m1 = 0, T = 0. Both of which are expected.

2 1

2 1

(m m )a g(m m )

−=

+

m1

m2

T

– T

m2g

– m1g

Notice that the acceleration of the two masses is the acceleration of gravity, g, times the difference in the masses divided by their sum.

The device thus effectively slows down the acceleration of the system due to gravity.

This is particularly useful if your time pieces are not very accurate as was the case in the 1780s when George Atwood invented the machine for more accurate measurement of g. If we solve for g:

2 1

2 1

(m m )g a(m m )

+=

−

m1

m2

From kinematics we have that,

2o o

1y y v t at2

= + +

a

m2

m1 Consider m1 starting from rest (vo=0) at the floor (yo=0) and rising to a height y in time t. Then,

21y at2

=

Solving for a,

2

2yat

=

Then, 2 1

22 1

(m m ) 2yg(m m ) t

+=

− Test by experiment

For a rope and pulley of negligible mass, no friction, m1 = 5m2 and θ = 30o, what is the magnitude and direction of the acceleration of the 2 blocks once released?

Example

Draw the free body diagram for each mass labeling all the forces that are acting on the mass. For m1 we resolve the force of gravity into components acting parallel and perpendicular to the incline.

Note that we did not put signs on the forces.

We don’t know in which direction the system will accelerate so we must make a guess. m1g

m1gsinθ

m1gcosθ m2g

T

T

θ

In the end if we guessed right the value of a will come out positive, if it’s actually in the opposite direction it will come out negative.

N

θ

If we chose down on the right and up the incline on the left for +a then for m1 along the direction parallel to the incline forces up the incline are positive and down are negative. So,

m1g m1gsinθ

m1gcosθ m2g

T

T

θ

a

a

1 1F m a=∑1 1T m gsin m a− θ =

For m2

2 2F m a=∑2 2m g T m a− =

Add the resulting eqns to eliminate T,

2 1 1 2m g m gsin m a m a− θ = +

2 1 1 2(m m sin )g (m m )a− θ = +

2 1

1 2

(m m sin )a g(m m )

− θ=

+

o(1 5sin30 )a g(5 1)

−=

+

Using that m1 = 5m2 and θ = 30o

o2 2

2 2

(m 5m sin30 )a g(5m m )−

=+

(1 2.5)a g6

−=

a 0.25g= −

So I guessed wrong and the system accelerates in the opposite direction.

m1g m1gsinθ

m1gcosθ m2g

T

T

θ

a

a

Friction

When a body in contact with a surface moves across the surface friction will induce a force in a direction to oppose the motion.

The reasons for this frictional force are complex: Simple mechanical interlocking plays a role. A degree of so call van der Walls bonding and potentially even some rapidly made and broken chemical bonds between atoms across the interface play a role.

The area of science that studies the causes and details of friction is called Tribology.

We will consider only the simplest model of friction.

Friction always acts as a retarding force (acts in a direction to oppose the motion) Its magnitude can often be modeled as,

f N= µ

Where f is the frictional force, N is the magnitude of the normal force, and µ is a coefficient that relates them.

mg

F N

f N= µ

Consider dragging a crate across the floor:

a

mg

N f 0=

If not for the friction with the floor, you’d need only apply a force to accelerate the crate and no further force would be needed to keep it going with whatever velocity you gave it.

mg

F

N f N= µv

Because of friction however, to keep the crate moving with constant velocity, v, you must continuously supply a force F, that precisely balances f.

v

(Since by the 2nd law F ma F f 0 F f= → − = → =∑0 for v constant

mg

F

N f N= µv

This provides a means for determining µ for the specific material of the crate and floor sliding against each other.

From the relation we have that , now since the f N= µfN

µ =

velocity is constant, F and f are equal but opposite so that |F| = |f|

FN

µ =So,

(constant v only)

mg

F

N f N= µv

f FN N

µ = = (constant v)

To get the block going faster we need merely supply force, F > |f| , to accelerate to the new speed, after which only the original force, F = |f| is needed to keep the crate going at the new speed.

If the applied force F drops below |f|, the crate comes to a stop.

Note that µ is an experimentally determined property of the particular materials sliding against one another.

If the crate and floor are made of wood, µ (wood-wood), will have a different value than if the crate has a Teflon skid, µ (Teflon-wood).

Example

Beatrice pulls the 36 kg suitcase at constant speed along the floor with a force of 70 N at the angle of 40o. What is µk between the suitcase and the floor?

θ

F

mg

f Fsinθ

Fcosθ

Ν

Free body diagram for suitcase:

With resolved along the x and y directions. F

2nd law for the x direction: xF Fcos f ma 0= θ − = =∑Fcos f 0θ − =

θ

F

mg

f Fsinθ

Fcosθ

Ν

We need to determine N

2nd law for the y direction: yF Fsin N mg ma 0= θ + − = =∑

f Fcos= θSo for the x direction we have:

But kf N= µ so k N Fcosµ = θ

Or, kFcos

Nθ

µ =

N mg Fsin= − θ

θ

F

mg

f Fsinθ

Fcosθ

Ν

We get

Then with & kFcos

Nθ

µ = N mg Fsin= − θ

kFcos

mg Fsinθ

µ =− θ

o

ko

2

(70N)cos(40 )m36 kg(9.8 ) (70N)sin(40 )s

µ =−

k 0.174µ =

If there is friction it is found that a minimum force must be supplied before the crate will move. The magnitude of this force has the same form as for kinetic friction . f N= µ

If the crate is initially at rest and there is no friction, the smallest of forces will accelerate it into motion.

So far we’ve considered the case of, , i.e. the case of kinetic friction.

v 0≠

It also depends on the normal force, and on a coefficient that depends on the surfaces in contact, but the value of the coefficient is different than in the kinetic case.

Kinetic friction: , coefficient of kinetic friction

Static friction: with , coefficient of static friction

HITT (survey)

An object is sitting on a horizontal surface at sea level with no external lateral force acting on it. Is there a force of friction, fs = µsN acting on it?

A = yes B = no

To distinguish the two cases we define:

k kf N= µ

s,max sf N= µ

kµ

sµsf

Answer: No

If you push on the object, the retarding force of static friction will (up to a maximum force) be precisely equal and opposite to the force you apply, keeping the object motionless. If there is no applied force, the force of friction is zero.

As your force increases, so will that of the static friction, until it reaches its maximum, when fs,max = µsN , at which point it breaks free and begins to accelerate (and kinetic friction takes over).

This is why is often written as an inequality.

Without an external force, in what direction would the static friction act?

s sf N≤ µ

Only at the instant just before an applied force is big enough to cause motion is,

s s,max sf f N= = µ (just before breaks free, v = 0) Example

The wood block of mass m = 2.0 kg is let go on a smooth steel ramp having an angle of θ = 25o to the horizontal.

For this material combination,

s 0.55µ = k 0.40µ =

(Note generally true is that , and that both are dimensionless) s kµ > µ

θ

Does the block slide and what is the force of friction?

mgθθ

F

F⊥

The force of gravity acting straight down on the block is resolved into components parallel and perpendicular to the incline.

The force of gravity is mg. From the right triangle drawn and trig,

adj Fcoshyp mg

⊥θ = =

Perpendicular component:

Giving F mgcos⊥ = θ

Foppsinhyp mg

θ = =

Parallel component:

Giving F mgsin= θ

mgθθ

N mgcos= θ

The force acting to slide the block down the incline is thus

The force that the block applies into incline is

The incline provides a normal force equal to this but in the opposite direction having magnitude.

F mgcos⊥ = θ

F mgsin= θ

F mgsin= θ

F mgcos⊥ = θN mgcos= θ

mgθθ

N mgcos= θ

Since the component of gravity parallel to the incline F|| acts in the direction down the incline, the force of friction, f, which always opposes the sliding will be up the incline.

F mgsin= θ

F mgcos⊥ = θ

f

Now if the block slides or not depends on whether or not

s,max sf N= µF|| exceeds

Check:

s,max s sf N mgcos= µ = µ θ

os,max 2

mf (0.55)(2kg)(9.8 )cos(25 ) 9.77Ns

= =

o2

mF mgsin (2kg)(9.8 )sin(25 ) 8.28Ns

= θ = =

Since F|| < fs,max the block does not move.

F

(not )

mgθθ

N mgcos= θ

Since the block does not move, Newton’s 2nd law requires that,

F mgsin= θ

F mgcos⊥ = θ

sf

F

F ma 0= =∑

sF f 0− =

sf F 8.28N= =

So the force of friction exactly balances F|| to make a = 0.

s,maxf 9.77N=

mgθ

θ

What happens if the angle of the incline increases?

F mgsin= θ

F mgcos⊥ = θ

sf

F

mgθθ

N mgcos= θ

F mgsin= θ

F mgcos⊥ = θ

sf

F

the force down the incline F|| grows while (and therefore N) decrease. At some angle the force down the incline will equal fs,max = µs N and for any greater angle the block will begin to slip.

Since sin θ grows with increasing θ while cos θ shrinks (0 < θ < 90o) F⊥

N mgcos= θ

mgθ

θ

At the angle right on the verge of slipping we have:

F mgsin= θ

F mgcos⊥ = θ

sf

F

F ma 0= =∑s,maxF f 0− =

s,maxF f=

max smgsin Nθ = µ

max s maxmgsin mgcosθ = µ θ

maxs

max

sincos

θ= µ

θ

max stan θ = µ 1 1 omax stan tan (0.55) 28.8− −→ θ = µ = =

N mgcos= θ

Suppose now that the angle of the incline is 32o.

or,

mgθ

θ

F mgsin= θ

F mgcos⊥ = θ

kf N= µ

F

N mgcos= θSince the block now slips the force of friction is given by, so now,

kf N= µ

F ma 0= ≠∑F f ma− =

kmgsin N maθ − µ =

kmgsin mgcos maθ − µ θ =

ka g(sin cos )= θ − µ θ

o o2 2

m ma (9.8 )(sin(32 ) (0.40)(cos(32 )) 1.87s s

= − =