PHY 231 1 PHYSICS 231 Lecture 24: Ideal gases Remco Zegers Walk-in hour:Tue 4-5 pm Helproom.
PHY 231 1 PHYSICS 231 Lecture 8: Forces, forces & examples Remco Zegers Walk-in hour: Monday...
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Transcript of PHY 231 1 PHYSICS 231 Lecture 8: Forces, forces & examples Remco Zegers Walk-in hour: Monday...
PHY 2311
PHYSICS 231Lecture 8: Forces, forces & examples
Remco ZegersWalk-in hour: Monday 11:30-13:30
Helproom
PHY 2312
Newton’s Laws
First Law: If the net force exerted on an object is zero the object continues in its original state of motion; if it was at rest, it remains at rest. If it was moving with a certain velocity, it will keep on moving with the same velocity.
Second Law: The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass: F=ma
If two objects interact, the force exerted by the first object on the second is equal but opposite in direction to the force exerted by the second object on the first: F12=-F21
PHY 2313
Forces seen in the previous lecture
• Gravity: Force between massive objects• Normal force: Elasticity force from supporting surface
Fg//=mgsinFgL=mgcos
Fg=mg
n=-FgL
PHY 2314
Gravity, mass and weights.
Weight=mass times gravitational accelerationFg(N)=M(kg) g(m/s2)
Newton’s law of universal gravitation:Fgravitation=Gm1m2/r2
G=6.67·10-11 Nm2/kg2
For objects on the surface of the earth:•m1=mearth=fixed
•r=“radius” of earth=fixed•The earth is not a point object relative to m2
PHY 2315
Measuring mass and weight.
Given that gearth=9.81 m/s2, gsun=274 m/s2, gmoon=1.67 m/s2,what is the mass of a person on the sun and moon if hismass on earth is 70 kg? And what is his weight on each ofthe three surfaces?
• The mass is the same on each of the surfaces
• On Earth: w=686.7 N• On the Moon: w=116.7 N• On the Sun: 19180 N
PHY 2316
Jumping!
The pelvis has a mass of 30.0 kg. Whatis its acceleration?
Decompose all forces in x and y directions
Force x (N) y (N)300 N -122 -274690 N 236 648Weight 0 -249Resultant 114 N 80.3 N
Total Force: F=(1142+80.32)=139 NDirection: =tan-1(Fy/Fx)=35.2o
Acceleration: a=F/m=139/30.0=4.65 m/s2
PHY 2317
TensionT The magnitude of the force T
acting on the crate, is the sameas the tension in the rope.
Spring-scale
You could measure the tension by insertinga spring-scale...
PHY 2318
Newton’s second law and tensionm1
m2
No friction.n
Fg
T
T
Fg
Object 1: F=m1a, so T=m1aObject 2: F=m2a, so Fg-T=m2a
m2g-T=m2aCombine 1&2 (Tension is the same): a=m2g/(m1+m2)
What is the acceleration ofthe objects?
PHY 2319
Problem
T
Draw the forces: what is positive &negative???
Fg
Fg
T
For 3.00 kg mass: F=ma T-9.813.00=3.00a
For 5.00 kg mass: F=ma 9.815.00-T=5.00a
What is the tension in the string andwhat will be the acceleration of thetwo masses?
T=36.8 Na=2.45 m/s2
PHY 23110
Friction
Friction are the forces acting on an object due to interaction
with the surroundings (air-friction, ground-friction etc).Two variants:•Static Friction: as long as an external force (F) trying to make an object move is smaller than fs,max, the static friction fs equals F but is pointing in the opposite direction: no movement!
fs,max=sn s=coefficient of static friction
•Kinetic Friction: After F has surpassed fs,max, the object starts moving but there is still friction. However, the friction will be less than fs,max!
fk=kn k=coefficient of kinetic friction
PHY 23111
PHY 23112
Problem
Fg//=mgsinFgL=mgcos
Fg=mg
n=-FgL
Fs,kA)If s=1.0, what isthe angle for which theblock just starts to slide?B)The block starts moving. Given that k=0.5, what is theacceleration of the block?
A) Parallel direction: mgsin-sn=0 (F=ma) Perpendicular direction: mgcos-n=0 so n=mgcos Combine: mgsin-smgcos=0 s=sin/cos=tan=1 so =45oB) Parallel direction: mgsin(45o)- smgcos(45o)=ma (F=ma) g(½2-¼2)=a so a=g¼2
PHY 23113
All the forces come together...
Fg
Tn
Fk
Fg
T
If a=3.30 m/s2 (the 12kg blockis moving downward), what isthe value of k?
For the 7 kg block parallel to the slope:T-mgsin-kmgcos=maFor the 12 kg block: Mg-T=Ma
25.0cos
sin)(
mg
mamgagMkSolve for k
PHY 23114
General strategy
•If not given, make a drawing of the problem.•Put all the relevant forces in the drawing, object by object.
• Think about the axis• Think about the signs
•Decompose the forces in direction parallel to the motion and perpendicular to it.
•Write down Newton’s first law for forces in the parallel direction and perpendicular direction.
•Solve for the unknowns.•Check whether your answer makes sense.
PHY 23115
Next Lecture:
Revision and go through an exam
May the Force be with you!