Newton's Third Law of Motion: Difference between revisions

Revision as of 16:09, 8 April 2017

By Karan Shah

Main Idea

Newton’s Third Law of Motion describes a push or pull that acts on an object as a result of its interaction with another object. According to this law for every action there is an equal and opposite re-action. This means that for every force there is a reaction force that is equal in size, but opposite in direction. Meaning that when an object 1 pushes another object 2 then object 1 gets pushed back with equal force but in the opposite direction.

If you push an object with 100N it will push back on you with equal but opposite force.

Mathematically Formula to describe Newton's Third Law

The third law of motion is also referred to as the action-reaction law because both objects are part of a single interaction and neither force can exist without the other. An important concept to remember about Newton's Third Law of Motion is that the two forces are of the same type. For example, when you throw a ball in the sky the Earth exerts a gravitational force on the ball and the ball also exerts a

The canon pushes the canon ball forward and the canon pushes the canon back with equal force.

gravitational force that is equal in magnitude and opposite in direction on the earth. Another example, that can sum up the concept of Newton's Third Law is when you walk. When you push down upon the ground and ground pushes with the same force upward. Similarly, the tires of a car push against the road while the road pushes back on the tires.

Examples

Here are some problems regarding Newton's Third Law.

Simple

Question

Car B is stopped at a red light. The brakes in Car A have failed and Car A is coming towards Car B at 60 kmh. Car B then runs into the back of Car A, What can be said about the force on Car A on Car B and the force on Car B on Car A?

Answer

B exerts the same amount of force on A as A exerts on B. 
Just the direction of both the forces will be in the opposite direction.

Middle

Question

Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed forward by a 8 N force applied to the 1 kg block. (a) How much force does the 2 kg block exert on the 3 kg block? (b) How much force does the 2 kg block exert on the 1 kg block?

Answer

(a) 
Find the Acceleration of the Whole Object:
Total Mass: 6kg
8 = (6) a 
a = 8 / 6 = 1.33 m/s^2
Total Acceleration: 1.33 m/s^2 (Acceleration will be the same for all three blocks)
F(2 on 3) = m(3) * a  
3 * 1.33 = 3.999 N

(b) 
Total Acceleration: 1.33 m/s^2
Mass to push: 5 kg (Because we are also pushing the 3 kg block)
F(1 on 2) = 5 * 1.33
F(1 on 2) = 5.33 N

Difficult

Question

A massive steel cable drags a 30 kg block across a horizontal, frictionless surface. A 100 N force applied to the cable causes the block to reach a speed of 5.0 m/s in a distance of 5.0 m. What is the mass of the cable?

Answer

a = V² / (2x) = 2.5 m/s² 
F = M * a → 100 = (30+m) * 2.5 
m = 10 kg

Real World Application

Scenario: There is a car that is stuck, nestled in the trees. Woman appears healthy and stable.

Witness Statement: Witness says that as a car was making a turn, the car “somehow lost its balance and skid”, deviating off its regular path, and ended up sliding towards the trees, around 30 metres from where she was. The witness claims the car was travelling at 25 miles per hour.

Report: This is a classic combination case of NEWTON’S 1st and 3rd LAW. The first law says that objects at a fixed velocity will stay at that velocity unless acted upon by an unbalanced, external force. It is worth pointing out that when the car was skidding, it was displaying inertia, or the tendency to resist a change in motion. In the car, as the wheels go around, it’s static friction. The force that is pushing the car along the road is a friction force... we need to the road to move the car, cause the road provides the friction. This is Newton’s 3rd Law. The tires are pushing backwards on the road, and the road is pushing forward on the tires. This “push force” is what we’re relying on keep the car on the road, so essentially, we’re relying on this static friction force. When the car starts to skid, it changes to force of kinetic friction, because the wheels are now slipping and having some motion. It slides and is moving, so it becomes kinetic friction.

The witness claimed that the car was going at 25 miles per hour. With this being said, an official wanted check and make sure this really was the speed. Since the car is now 30 metres from the place it lost its balance, we can use this information to calculate the speed. Assuming the brakes work properly, the stopping distance would’ve been determined by the coefficient of friction between the tires and the road. This is because the force of friction must do enough work on the car to reduce the kinetic energy of the car, as outlined in the work-energy principle. Net Work = change in kinetic energy of an object Net Work = (1/2)mv(final)^2 - (1/2)mv(initial)^2 When we look at the equation for static friction: Ms = Ff / Fn Fn x Ms = Ff Ms x mg = Fr Mu static x mg = Force of friction Since the road was a bit slippery, and the road a little wet, the coefficient of static friction was 0.4. The normal coefficient is around 0.7 for dry roads and becomes smaller for wet or oily roads. Gravity was 10, so the equation would’ve been:

Work Friction = -μmgd = - (1/2) mv^2

If we cancel out the m’s, we can get D = (v^2) / (2)(μ)(g) Plugging in what we have and what we know: 30 = (V^2) / 2 x .4 x 10 30 = (V^2) / 8 V = 15 metre / second, so around 35 miles per hour. The witness’ account of the speed was a little off!

Conclusion: Like most skids, as soon as this car started to skid, it was at the mercy of inertia. It would keep going at the same speed unless some other force stopped it, and in this case, that “other force” was the tress.

Connectedness

How is this topic connected to something that you are interested in?

Newton's Third Law is connected to the concept of a spacecraft flying in space. When a spacecraft 
fires a thruster rocket, the exhaust gas pushes against the thruster and the thruster pushes 
against the exhaust gas. The gas and rocket move in opposite directions. This is an 
example of Newton's Third Law because both forces are equal in magnitude and opposite 
in direction. This topic is related to space travel a topic that I am interested in and 
passionate about learning more.

History

Sir Isaac Newton was a renowned scientist and mathematician who helped create a foundation for modern studies. He was born in England in 1643 and worked his way to earn a bachelor’s and master’s degree from Trinity College Cambridge. He was highly interested in math, physics, and astronomy and wrote many of his ideas in a journal. One of those ideas was about the three laws of motion.

In 1687 Isaac Newton made his work on his book, Philosophiae Naturalis Principia Mathematic or Principia known to the public. He discussed the principles of time, force, and motion that helped create modern physical science and helped account for much of the phenomena viewed in the world. Some of the principles he discusses include acceleration, initial movement, fluid dynamics, and motion. Newton’s Laws first appeared in the Principia and discussed the relationship that exists between forces acting on a body and the motion of the body. For the third law, he stated that for every action/force in nature, there will be an equal and opposite reaction.

http://www.livescience.com/46561-newton-third-law.html

Newton's Third Law of Motion: Difference between revisions

Revision as of 16:09, 8 April 2017

Contents

Main Idea

Examples

Simple

Question

Answer

Middle

Question

Answer

Difficult

Question

Answer

Real World Application

Connectedness

History

See also

Further reading

External links

References

Navigation menu

@@ Line 59: / Line 59: @@
   F = M * a → 100 = (30+m) * 2.5
   m = 10 kg
+===Real World Application===
+'''Scenario''': There is a car that is stuck, nestled in the trees. Woman appears healthy and stable.
+'''Witness Statement''': Witness says that as a car was making a turn, the car “somehow lost its balance and skid”, deviating off its regular path, and ended up sliding towards the trees, around 30 metres from where she was. The witness claims the car was travelling at 25 miles per hour.
+'''Report''': This is a classic combination case of NEWTON’S 1st and 3rd LAW. The first law says that objects at a fixed velocity will stay at that velocity unless acted upon by an unbalanced, external force. It is worth pointing out that when the car was skidding, it was displaying inertia, or the tendency to resist a change in motion. In the car, as the wheels go around, it’s static friction. The force that is pushing the car along the road is a friction force... we need to the road to move the car, cause the road provides the friction. This is Newton’s 3rd Law. The tires are pushing backwards on the road, and the road is pushing forward on the tires. This “push force” is what we’re relying on keep the car on the road, so essentially, we’re relying on this static friction force. When the car starts to skid, it changes to force of kinetic friction, because the wheels are now slipping and having some motion. It slides and is moving, so it becomes kinetic friction.
+The witness claimed that the car was going at 25 miles per hour. With this being said, an official wanted check and make sure this really was the speed. Since the car is now 30 metres from the place it lost its balance, we can use this information to calculate the speed.
+Assuming the brakes work properly, the stopping distance would’ve been determined by the coefficient of friction between the tires and the road. This is because the force of friction must do enough work on the car to reduce the kinetic energy of the car, as outlined in the work-energy principle.
+Net Work = change in kinetic energy of an object
+Net Work = (1/2)mv(final)^2 - (1/2)mv(initial)^2
+When we look at the equation for static friction:
+Ms = Ff / Fn
+Fn x Ms = Ff
+Ms x mg = Fr
+Mu static x mg = Force of friction
+Since the road was a bit slippery, and the road a little wet, the coefficient of static friction was 0.4. The normal coefficient is around 0.7 for dry roads and becomes smaller for wet or oily roads. Gravity was 10, so the equation would’ve been:
+ Work Friction = -μmgd = - (1/2) mv^2
+If we cancel out the m’s, we can get
+D = (v^2) / (2)(μ)(g)
+Plugging in what we have and what we know:
+= (V^2) / 2 x .4 x 10
+= (V^2) / 8
+V = 15 metre / second, so around 35 miles per hour. The witness’ account of the speed was a little off!
+Conclusion: Like most skids, as soon as this car started to skid, it was at the mercy of inertia. It would keep going at the same speed unless some other force stopped it, and in this case, that “other force” was the tress.
 ==Connectedness==