# The 3 rd 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?

## Presentation on theme: "The 3 rd 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?"— Presentation transcript:

The 3 rd 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

F gases F forward

ICE

A 200 kg crate is loaded onto a 2000 kg truck. The force that propels the truck can be written as F T. Calculate the maximum magnitude of F T 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/s 2 ) b)Calculate the tension in the cord between the boxes. (21.8 N) 10.0 kg12.0 kg F p = 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 = m 1 a 1 = F T – m 1 g  F = m 2 a 2 = F T – m 2 g Or m 1 a 1 = F T – m 1 g m 2 a 2 = F T – m 2 g We have two equations, but three unknowns (F T, a 1, and a 2 )

However, since the elevator will drop the counterweight will rise: a 1 = -a 2 m 1 a 1 = F T – m 1 g m 2 a 2 = F T – m 2 g Three equations, three unknowns (F T, a 1, and a 2 )

m 1 a 1 = F T – m 1 gSubstitute a 1 = -a 2 -m 1 a 2 = F T – m 1 gSolve for F T F T = m 1 g - m 1 a 2 m 2 a 2 = F T – m 2 gSubstitute for F T m 2 a 2 = m 1 g - m 1 a 2 – m 2 g m 2 a 2 + m 1 a 2 = m 1 g– m 2 g a 2 (m 2 +m 1 )=g(m 1 -m 2 ) a 2 = g(m 1 -m 2 )= 9.8(1150-1000) = 0.68 m/s 2 (m 2 +m 1 ) (1000+1150)

Since we have solved for a 2, it makes most sense to use this equation to find F T m 2 a 2 = F T – m 2 g F T = m 2 a 2 + m 2 g F T = (1000 kg)(0.68m/s 2 ) +(1000 kg)(9.8 m/s 2 ) F T =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 = 2F T –mg = maa = 0 (constant vel.) 0 = 2F T –mg Rearrange F T = mg/2Substitute F T =(200kg)(9.8m/s 2 )/2 F T =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 (F push =300 N). If the care begins to budge at an angle of 5 o, 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.

 F x = 0 = F T1x – F T2x  F y = 0 = F p - F T1y – F T2y 0 = F T1x – F T2x 0 = F p - F T1y – F T2y Use trigonometry 0 = F T1 cos5 o – F T2 cos 5 0 0 = 300N - F T1 sin5 o – F T2 sin5 o

F T1 cos5 o = F T2 cos 5 o Rearrange F T1 cos5 o = F T2 cos 5 o F T1 = F T2 Substitute 0 = 300N – F T2 sin5 o – F T2 sin5 o 300N = 2F T2 sin5 o F T2 = 300N/2sin5 o = 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/s 2 ) b.Calculate the tension (17 N) m 1 =5.0 kg m 2 =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 . (30 o ) 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/s 2 ) 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/s 2 to shove it up a 30 o 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)

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