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