SPH4U: Practice ProblemsSPH4U: Practice Problems

Today’s AgendaToday’s Agenda

Run and Hide

Review: Pegs & PulleysReview: Pegs & Pulleys Used to change the direction of forcesUsed to change the direction of forces

An ideal massless pulley or ideal smooth peg will change the An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: direction of an applied force without altering the magnitude: The The tension is the same on both sides!tension is the same on both sides!

FF1 = -T ii ideal peg

or pulley

FF2 = T jj

| FF1 | = | FF2 |

massless rope

Problem: AccelerometerProblem: Accelerometer A weight of mass A weight of mass mm is hung from the ceiling of a car with a massless is hung from the ceiling of a car with a massless

string. The car travels on a horizontal road, and has an acceleration string. The car travels on a horizontal road, and has an acceleration aa in the in the xx direction. The string makes an angle direction. The string makes an angle with respect to the with respect to the vertical (vertical (yy) axis. Solve for ) axis. Solve for in terms of in terms of aa and and gg..

a

i i

Accelerometer...Accelerometer... Draw a Draw a free body diagramfree body diagram for the mass: for the mass:

What are all of the forces acting?What are all of the forces acting?

m

TT (string tension)

mgg (gravitational force)

i i

Accelerometer...Accelerometer...

Using components Using components (recommended):(recommended):

ii: : FFX X = T= TXX = T sin = T sin = ma= ma

jj:: F FY Y = T= TY Y mg mg

= T cos = T cos mg = 0 mg = 0

TT

mgg

mmaa

jj

ii

TX

TY

Accelerometer...Accelerometer...

Using components Using components ::

ii: : T sin T sin = = mama

jj:: T cos T cos - mg = 0- mg = 0

Eliminate Eliminate T T :: mgg

m

maa

T sin = ma

T cos = mgtan

a

g

TX

TY

jj

ii

TT

Accelerometer...Accelerometer... Alternative solution using vectorsAlternative solution using vectors (elegant but not as systematic): (elegant but not as systematic):

Find the total vector force Find the total vector force FFNETNET::

TT

mgg

FFTOT

m

TT (string tension)

mgg (gravitational force)

Accelerometer...Accelerometer... Alternative solution using vectorsAlternative solution using vectors (elegant but not as systematic): (elegant but not as systematic): Find the total vector force Find the total vector force FFNETNET::

Recall that Recall that FFNET NET = m= maa::

So So

maa

tanma a

mg g

TT

mgg

tana

g

m

TT (string tension)

mgg (gravitational force)

Accelerometer...Accelerometer... Let’s put in some numbers:Let’s put in some numbers:

Say the car goes from Say the car goes from 00 to to 60 mph60 mph in in 10 seconds10 seconds:: 60 mph = 60 x 0.45 m/s = 27 m/s60 mph = 60 x 0.45 m/s = 27 m/s.. Acceleration Acceleration a = Δv/Δt = 2.7 m/sa = Δv/Δt = 2.7 m/s22.. So So a/g = 2.7 / 9.8 = 0.28 a/g = 2.7 / 9.8 = 0.28 ..

= arctan= arctan (a/g) = 15.6 deg(a/g) = 15.6 deg

tana

g

a

UnderstandingUnderstanding

A person standing on a horizontal floor feels two forces: the downward pull of gravity and the upward supporting force from the floor. These two forces

(A) Have equal magnitudes and form an action/reaction pair(B) Have equal magnitudes but do not form an action/reaction pair(C) Have unequal magnitudes and form an action/reaction pair(D) Have unequal magnitudes and do not form an action/reaction

pair(E) None of the aboveBecause the person is not accelerating, the net force they feel is zero. Therefore the magnitudes must be the same (opposite directions. These are not action/reaction forces because they act of the same object (the person). Action/Reaction pairs always act on different objects.

Angles of an Inclined planeAngles of an Inclined plane

ma = mg sin

mgN

The triangles are similar, so the angles are the same!

Problem: Inclined planeProblem: Inclined plane

A block of massA block of mass mm slides down a frictionless ramp that makes angleslides down a frictionless ramp that makes angle with respect to the horizontal. What is its acceleration with respect to the horizontal. What is its acceleration aa ??

ma

Inclined plane...Inclined plane... Define convenient axes parallel and perpendicular to plane:Define convenient axes parallel and perpendicular to plane:

Acceleration Acceleration aa is in is in xx direction only. direction only.

ma

ii

jj

Inclined plane...Inclined plane... Consider Consider xx and and yy components separately:components separately: ii: : mg sin mg sin = =mama. . a = g sin a = g sin

jj: : N - mg cos N - mg cos = 0= 0. . N = mg cos N = mg cos

mgg

NN

mg sin

mg cos

maa

ii

jj

Problem: Two Blocks Problem: Two Blocks

Two blocks of masses Two blocks of masses mm11 and and mm22 are placed in contact on a are placed in contact on a

horizontal frictionless surface. If a force of magnitude horizontal frictionless surface. If a force of magnitude FF is applied to is applied to the box of mass the box of mass mm11, what is the force on the block of mass , what is the force on the block of mass mm22??

mm11 mm22

F

Problem: Two BlocksProblem: Two Blocks Realize that Realize that FF == ((mm11+ + mm22)) aa ::

Draw FBD of block Draw FBD of block mm22 and apply and apply FFNET NET = m= maa::

F2,1F2,1 = mm2 2 a

F / (mm11+ mm22) = a

m22,1

m2m1

FF

Substitute for a :

mm22

(m1 + m2)m2F2,1 F

Problem: Tension and AnglesProblem: Tension and Angles

A box is suspended from the ceiling by two ropes making an angle A box is suspended from the ceiling by two ropes making an angle with the horizontal. What is the tension in each rope?with the horizontal. What is the tension in each rope?

mm

Problem: Tension and AnglesProblem: Tension and Angles

Draw a FBD:Draw a FBD:

T1 T2

mg

Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0

T1sin T2sin

T2cos T1cos

jj

ii

Fx,NET = T1cos - T2cos = 0 T1 = T2

2 sin mg

T1 = T2 =Fy,NET = T1sin + T2sin - mg = 0

Problem: Motion in a CircleProblem: Motion in a Circle

A boy ties a rock of mass A boy ties a rock of mass mm to the end of a string and twirls to the end of a string and twirls it in the vertical plane. The distance from his hand to the it in the vertical plane. The distance from his hand to the rock is rock is RR. The speed of the rock at the top of its trajectory is . The speed of the rock at the top of its trajectory is vv.. What is the tension What is the tension TT in the string at the top of the rock’s in the string at the top of the rock’s

trajectory?trajectory?

R

v

TT

Motion in a Circle...Motion in a Circle... Draw a Free Body Diagram (pick Draw a Free Body Diagram (pick yy-direction to be down):-direction to be down): We will use We will use FFNETNET = m = maa (surprise)(surprise)

First find First find FFNET NET in in yy direction: direction:

FFNETNET = mg +T = mg +T

TTmgg

y y

Motion in a Circle...Motion in a Circle...

FFNETNET = = mgmg ++TT

Acceleration inAcceleration in y y direction: direction:

mmaa = mv= mv22 / R / R

mg + mg + TT = mv = mv22 / R / R

T T = mv= mv22 / R - mg / R - mg

R

TT

v

mgg

y y

F = ma

Motion in a Circle...Motion in a Circle... What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp?What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp?

i.e. find i.e. find vv such that such that T = 0T = 0..

mvmv22 / R = mg + T / R = mg + T

vv22 / R = g / R = g

Notice that this doesNotice that this doesnotnot depend on depend on mm ..

R

mgg

v

T= 0

v Rg

UnderstandingUnderstandingTwo-body dynamicsTwo-body dynamics

In which case does block In which case does block m m experience a larger acceleration? Inexperience a larger acceleration? In case (1) case (1) there is a there is a 10 kg10 kg mass hanging mass hanging from a rope. In from a rope. In case (2) case (2) a hand is providing a constant downward force of a hand is providing a constant downward force of 98.1 N98.1 N. . In both cases the ropes In both cases the ropes and pulleys are massless.and pulleys are massless.

(a)(a) Case (1) (b) (b) Case (2) (c)(c) same

m

10kga a

m

F = 98.1 N

Case (1) Case (2)

SolutionSolution

m

10kga

Add (a) and (b):

mWg = (m + mW)a

W

W

m ga

m m

(a)

(b)

T = ma (a)

mWg -T = mWa (b)

For case (1) draw FBD and write FNET = ma for each block:

mW=10kg

SolutionSolution

The answer is (b) Case (2). In this case the block experiences a larger acceleration

98.1Na

mT = 98.1 N = ma For case (2)

m

10kga

Case (1)

kg10m

N198a

.

m

a

F = 98.1 N

Case (2)

m

N198a

.

Problem: Two strings & Two Masses onProblem: Two strings & Two Masses onhorizontal frictionless floor:horizontal frictionless floor:

m2 m1T2 T1

GivenGiven TT11, , mm11 and and mm22, what are , what are aa and and TT22??

TT1 1 - T- T22 = = mm11aa (a)(a)

TT2 2 = = mm22aa (b) (b)

Add (Add (a)a) + + (b)(b)::

TT1 1 = (m= (m1 1 + m+ m22)a)a a a

a

i i

1

1 2

T

m m

Plugging solution into (b): 22 1

1 2

mT T

m m

UnderstandingUnderstandingTwo-body dynamicsTwo-body dynamics

Three blocks of mass Three blocks of mass 3m3m, , 2m2m, and , and mm are connected by strings and pulled with constant are connected by strings and pulled with constant acceleration acceleration aa. What is the relationship between the tension in each of the strings? . What is the relationship between the tension in each of the strings?

(a) T1 > T2 > T3 (b) T3 > T2 > T1 (c) T1 = T2 = T3

T3 T2 T13m 2m m

a

SolutionSolution

Draw free body diagrams!!Draw free body diagrams!!

T33mT3 = 3ma

T3 T22mT2 - T3 = 2ma

T2 = 2ma +T3 > T3

T2 T1m

T1 - T2 = ma

T1 = ma + T2 > T2

T1 > T2 > T3

Alternative Alternative SolutionSolution

T3 T2 T13m 2m m

a

Consider T1 to be pulling all the boxes

T3 T2 T13m 2m m

a

T2 is pulling only the boxes of mass 3m and 2m

T3 T2 T13m 2m m

a

T3 is pulling only the box of mass 3m

T1 > T2 > T3

Problem: Rotating puck & weight.Problem: Rotating puck & weight.

A mass A mass mm11 slides in a circular path with speed slides in a circular path with speed vv on a on a

horizontal frictionless table. It is held at a radius horizontal frictionless table. It is held at a radius RR by by a string threaded through a frictionless hole at the a string threaded through a frictionless hole at the center of the table. At the other end of the string center of the table. At the other end of the string hangs a second mass hangs a second mass mm22.. What is the tension (What is the tension (TT) in the string?) in the string? What is the speed (What is the speed (vv) of the sliding mass?) of the sliding mass?

m1

m2

v

R

Problem: Rotating puck & weight...Problem: Rotating puck & weight...

DrawDraw FBDFBD of hanging mass:of hanging mass: SinceSince RR is constantis constant, , a = 0a = 0..

soso T = mT = m22gg

m2

m2g

T

m1

m2

v

R

T

Problem: Rotating puck & weight...Problem: Rotating puck & weight...

DrawDraw FBDFBD of sliding mass:of sliding mass:

m1

T = m2g

2

1

mv gR

m

m1g

N

m1

m2

v

R

T

Use F = T = m1a

where a = v2 / R

m2g = m1v2 / R

T = m2g