The 3rd Law: “For every force, there is an equal and opposite force.”

Runner example:• Does the runner push on the earth?• Why does the runner move more?• Does the earth move at all?

Forward force for the runnerBackward force for the earth

Fgases

Fforward

ICE

A 200 kg crate is loaded onto a 2000 kg truck. The force that propels the truck can be written as FT. Calculate the maximum magnitude of FT that can be applied before the crate slips (s = 0.80). Be sure to consider the force of the crate on the truck in your calculations.

(17,000N)

Tension

A cord is connected to a wall and a person pulls to the left with a force of 100 N.

A different cord is used in tug of war. Two people pull with 100 N force each, and the cord does not move.

Comment on the tension on the cord in both cases.

Two boxes are connected by a cord as shown. They are then pulled by another short cord.

a) Find the acceleration of each box (1.82 m/s2 )b) Calculate the tension in the cord between the boxes.

(21.8 N)

10.0 kg12.0 kg

Fp= 40.0 N

Tension: Example 2

Calculate the acceleration of the elevator and the tension in the cable.

Draw free-body diagrams for both the elevator and counterweight

Set up the force equations:F = m1a1 = FT – m1g

F = m2a2 = FT – m2g

Orm1a1 = FT – m1g

m2a2 = FT – m2g

We have two equations, but three unknowns (FT, a1, and a2)

However, since the elevator will drop the counterweight will rise:

a1 = -a2

m1a1 = FT – m1g

m2a2 = FT – m2g

Three equations, three unknowns (FT, a1, and a2)

m1a1 = FT – m1g Substitute a1 = -a2

-m1a2 = FT – m1g Solve for FT

FT = m1g - m1a2

m2a2 = FT – m2g Substitute for FT

m2a2 = m1g - m1a2 – m2g

m2a2 + m1a2 = m1g– m2g

a2(m2+m1)=g(m1-m2)

a2 = g(m1-m2) = 9.8(1150-1000) = 0.68 m/s2

(m2+m1) (1000+1150)

Since we have solved for a2, it makes most sense to use this equation to find FT

m2a2 = FT – m2g

FT = m2a2 + m2g

FT = (1000 kg)(0.68m/s2) +(1000 kg)(9.8 m/s2)

FT =10500 N

Mr. Fredericks uses a pulley to lift a 200 kg piano at a constant velocity. How much tension does he need to put on the rope?

F = 2FT –mg = ma a = 0 (constant vel.)

0 = 2FT –mg Rearrange

FT = mg/2 Substitute

FT =(200kg)(9.8m/s2)/2

FT =980 N (Note how the pulley

doubles my effort force.)

A physics student gets stuck in the mud. In order to get out, she ties a rope to a tree and pushes at the midpoint (Fpush=300 N). If the care begins to budge at an angle of 5o, calculate the force of the rope pulling on the car.

Note that the tension is always along the direction of the rope, and provided by the tree and the car. Since the car is just starting to budge, we will assume the sum of all the Forces is zero.

Fx = 0 = FT1x – FT2x

Fy = 0 = Fp - FT1y – FT2y

0 = FT1x – FT2x

0 = Fp - FT1y – FT2y Use trigonometry

0 = FT1cos5o – FT2cos 50

0 = 300N - FT1sin5o – FT2sin5o

FT1cos5o = FT2cos 5o Rearrange

FT1cos5o = FT2cos 5o

FT1 = FT2 Substitute

0 = 300N – FT2sin5o – FT2sin5o

300N = 2FT2sin5o

FT2 = 300N/2sin5o = 1700 N

(Note that she magnified her force almost 6 times!!!!)

In the following setup, the coefficient of kinetic friction between the box and the table is 0.20.

a. Calculate the acceleration of the system. (1.4 m/s2)

b.Calculate the tension (17 N)

m1=5.0 kg

m2=2.0 kg

A 90.0 kg mountain climber climbs from the ropes as shown. The maximum tension that rope 3 can hold is 1500 N before it breaks. Calculate the maximum angle of . (30o)

Rope 3

Rope 2

Rope 1

A 200 kg stage set is lifted down by a 100 kg stagehand as shown. Calculate the stagehand’s acceleration. (3.27 m/s2)

200 kg

100 kg

A 40 kg boy is working at his father’s store. He needs to give a 15 kg package an acceleration of 1.0 m/s2 to shove it up a 30o ramp. The coefficient of friction between the package and the ramp is 0.40. The coefficient between the boy and the slippery floor is only 0.25.

a. Calculate the force of the shove the boy must give to the package (140 N)

b. Calculate the normal force on the boy (462 N)c. Calculate whether the boy can provide enough

force considering the friction (no, only 115 N)