Solving Problems with Newton’s Laws “It sounds like an implosion!”
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Transcript of Solving Problems with Newton’s Laws “It sounds like an implosion!”
Solving Problems with Newton’s Laws
“It sounds like an implosion!”
Note that forces are VECTORS!!Newton’s 2nd Law: ∑F = ma
∑F = VECTOR SUM of all forces on mass m
We need VECTOR addition to add forces in the 2nd Law!
Forces add according to the rules ofVECTOR ADDITION!
(next chapter)• In this chapter, we consider only 1
dimensional motion & therefore only 1 dimensional Force vectors
Problem Solving Procedures1. Make a sketch. For each object
separately, sketch a free-body diagram, showing all forces acting on that object. Make the magnitudes & directions as accurate as you can. Label each force.
2. Apply Newton’s 2nd Law separately to each object.
3. Solve for the unknowns. Note that this often requires algebra, such
as solving 2 linear equations in 2 unknowns!
Applications & Examples of Newton’s 2nd Law in 1 Dimensional Motion
Transmitting Forces• Strings exert a force on the objects
they are connected to–This also applies to cables, ropes, etc.
• The mass of the cable may have to be taken into account
• Pulleys can redirect forces
• Forces can be amplified
Tension• Strings exert a force on the
objects they are connected to– Cables & ropes act the
same way• The strings exert force
due to their tension• The ends of the string
both exert a force of magnitude T on the supports where they are connected.
T is the tension in the string.
Tension Example – Elevator Cable• Two forces are acting
on the compartment– Gravity acting downward– Tension in cable acting
upward, T• Assume an acceleration
upward• Applying Newton’s
Second Law gives
T- mg = ma
• Now consider the cable– Assume the cable is massless
• Applying Newton’s 2nd Law gives: TC = T
• Tension is the same for all points along the cable
• True for all massless cables
• Tension has force units
Cables with Mass• Apply Newton’s 2nd Law
to the cable• To support the cable, the upper
tension, T1 must be larger than the tension from the box, T2
• If there is no acceleration, Newton’s 2nd Law is
T1 -T2 - mcable g = 0• We can assume a massless
cable if the mass of the cable is small compared to the other masses in the problem.
Single Pulleys• We often need to change
the direction of the force.• A simple pulley changes
the direction of the force, but not the magnitude– See diagram– Assume the rope and
pulley are both massless– Assume the cable does
not slip on the pulley
Pulleys To Amplify Forces• The person exerts a force T on
the rope.• The rope exerts a force 2T on
the pulley.• This force can be used to lift an
object.• More complex sets of pulleys
can amplify an applied force by greater factors.– The distance decreases to
compensate for the increase in force
Example (“Atwood’s Machine”)Two masses suspended over a (massless frictionless) pulley by a flexible (massless) cable “Atwood’s machine”. Example: Elevator & counterweight. Figure: Counterweight mC = 1000 kg. Elevator mE = 1150 kg. Calculate a) Elevator’s acceleration. b) Tension in the cable.
aE = - aaC = a
a
a
Free Body Diagrams
General Approach to Problem Solving1.Read the problem carefully; then read it again.
2.Draw a sketch, then a free-body diagram.
3.Choose a convenient coordinate system.
4.List the known & unknown quantities; find relationships between the knowns & the unknowns.
5.Estimate the answer.
6.Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in.
7.Keep track of dimensions.
8.Make sure your answer is REASONABLE!
Two boxes connected by a lightweight (massless!) cord are resting on a smooth (frictionless!) table. mA = 10 kg & mB = 12 kg. A horizontal force FP = 40 N is applied to mA. Calculate: a. Acceleration of the boxes. b. Tension in cord connecting the boxes.
Example
Free Body Diagrams
Problem FT1
m1g FT2
FT2
a
m2g
m1 = m2 = 3.2 kg, m1g = m2g = 31.4 NAcceleration a = 2.0 m/s2
Calculate FT1 & FT2
Use Newton’s 2nd Law: ∑Fy = mafor EACH bucket separately!!!
Take up as positive. Bucket 1: FT1 - FT2 - m1g = m1a (1)
Bucket 2: FT2 - m2g = m2a (2)
From (2), FT2 = m2(g + a) = (3.2)(9.8 + 2.0)
or FT2 = 37.76 NPut this into (1)
FT1 - m2(g + a) - m1g = m1a
Gives: FT1 = m2(g + a) + m1(g+a)
or FT1 = (m2+m1) (g + a) = 75.5 N
a
ProblemAcceleration a = 2.0 m/s2
m = 65 kg, mg = 637 NCalculate FT & FP
Take up as positive.Newton’s 2nd Law: ∑F = ma
(y direction) on woman + bucket!
FT + FT - mg = ma2FT - mg = ma
FT = (½)m(g + a) = 383.5 NAlso, Newton’s 3rd Law says that
FP = - FT = -383.5 N
FT
a
mg
FP
FT
a