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This lesson submitted by amanda.bragg

**Newton's Second Law of Motion**

According to Newton I, things 'like to keep on doing what they are doing'. In other words, if an object is moving, it tends to continue moving (in a straight line and at the same speed) and if an object is stationary, it tends to remain stationary. So how do objects start moving?

Let us look at the example of a 10 kg box on a rough table. If we push lightly on the box as indicated in the diagram, the box won't move. Let's say we applied a force of 100 N, yet the box remains stationary. At this point a frictional force of 100 N is acting on the box, preventing the box from moving. If we increase the force, let's say to 150 N and the box almost starts to move, the frictional force is 150 N. To be able to move the box, we need to push hard enough to overcome the friction and then move the box. If we therefore apply a force of 200 N remembering that a frictional force of 150 N is present, the 'first' 150 N will be used to overcome or 'cancel' the friction and the other 50 N will be used to move (accelerate) the block. In order to accelerate an object we must have a resultant force acting on the block.

http://cnx.org/content/m38963/latest/PG11C11_019.png

Now, what do you think will happen if we pushed harder, lets say 300 N? Or, what do you think will happen if the mass of the block was more, say 20 kg, or what if it was less? Let us investigate how the motion of an object is affected by mass and force.

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**Investigation : Newton's Second Law of Motion**

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**Aim:**

To investigate the relation between the acceleration of objects and the application of a constant resultant force.

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**Method:**

http://cnx.org/content/m38963/latest/PG11C11_020.png

- A constant force of 20 N, acting at an angle of 60∘ to the horizontal, is applied to a dynamics trolley.
- Ticker tape attached to the trolley runs through a ticker timer of frequency 20 Hz as the trolley is moving on the frictionless surface.
- The above procedure is repeated 4 times, each time using the same force, but varying the mass of the trolley as follows:

- Case 1: 6,25 kg
- Case 2: 3,57 kg
- Case 3: 2,27 kg
- Case 4: 1,67 kg

http://cnx.org/content/m38963/latest/PG11C11_021.png

Instructions:

- Use each tape to calculate the instantaneous velocity (in m⋅s−1) of the trolley at points B and F (remember to convert the distances to m first!). Use these velocities to calculate the trolley's acceleration in each case.
- Tabulate the mass and corresponding acceleration values as calculated in each case. Ensure that each column and row in your table is appropriately labeled.
- Draw a graph of acceleration vs. mass, using a scale of 1 cm = 1 m⋅s−2 on the y-axis and 1 cm = 1 kg on the x-axis.
- Use your graph to read off the acceleration of the trolley if its mass is 5 kg.
- Write down a conclusion for the experiment.

You will have noted in the investigation above that the heavier the trolley is, the slower it moved. The acceleration is *inversely* proportional to the mass. In mathematical terms:

*a*∝1/*m*

In a similar investigation where the mass is kept constant, but the applied force is varied, you will find that the bigger the force is, the faster the object will move. The acceleration of the trolley is therefore *directly* proportional to the resultant force. In mathematical terms:

*a*∝*F*

Rearranging the above equations, we get *a*∝*F/m* OR *F*=*ma*

Newton formulated his second law as follows:

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**DEFINITION 1: Newton's Second Law of Motion**

If a resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant force. The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the body. The mathematical representation is:

*F*=*ma*.

http://www.youtube.com/watch?feature=player_embedded&v=ou9YMWlJgkE

**Applying Newton's Second Law**

Newton's Second Law can be applied to a variety of situations. We will look at the main types of examples that you need to study.

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**EXERCISE 1: Newton II - Box on a surface 1**

A 10 kg box is placed on a table. A horizontal force of 32 N is applied to the box. A frictional force of 7 N is present between the surface and the box.

- Draw a force diagram indicating all the horizontal forces acting on the box.
- Calculate the acceleration of the box.

http://cnx.org/content/m38963/latest/PG11C11_022.png

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**EXERCISE 2: Newton II - box on surface 2**

Two crates, 10 kg and 15 kg respectively, are connected with a thick rope according to the diagram. A force of 500 N is applied. The boxes move with an acceleration of 2 m⋅s−2. One third of the total frictional force is acting on the 10 kg block and two thirds on the 15 kg block. Calculate:

- the magnitude and direction of the frictional force present.
- the magnitude of the tension in the rope at T.

http://cnx.org/content/m38963/latest/PG11C11_024.png

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**EXERCISE 3: Newton II - Man pulling a box**

A man is pulling a 20 kg box with a rope that makes an angle of 60∘ with the horizontal. If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.

http://cnx.org/content/m38963/latest/PG11C11_027.png

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**EXERCISE 4: Newton II - Truck and trailor**

A 2000 kg truck pulls a 500 kg trailer with a constant acceleration. The engine of the truck produces a thrust of 10 000 N. Ignore the effect of friction.

- Calculate the acceleration of the truck.
- Calculate the tension in the tow bar T between the truck and the trailer, if the tow bar makes an angle of 25∘ with the horizontal.

http://cnx.org/content/m38963/latest/PG11C11_029.png

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**Object on an inclined plane**

When we place an object on a slope the force of gravity (F*g*) acts straight down and not perpendicular to the slope. Due to gravity pulling straight down, the object will tend to slide down the slope with a force equal to the horizontal component of the force of gravity (F*g* sin *θ*). The object will 'stick' to the slope due to the frictional force between the object and the surface. As you increase the angle of the slope, the horizontal component will also increase until the frictional force is overcome and the object starts to slide down the slope.

The force of gravity will also tend to push an object 'into' the slope. The vertical component of this force is equal to the vertical component of the force of gravity (F*g* cos *θ*). There is no movement in this direction as this force is balanced by the slope pushing up against the object. This “pushing force” is called the normal force (N) and is equal to the force required to make the component of the resultant force perpendicularly into the plane zero, F*g* cos *θ* in this case, but opposite in direction.

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**TIP: **

Do not use the abbreviation W for weight as it is used to abbreviate 'work'. Rather use the force of gravity F*g* for weight.

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**EXERCISE 5: Newton II - Box on inclined plane**

A body of mass M is at rest on an inclined plane.

http://cnx.org/content/m38963/latest/PG11C11_033.png

What is the magnitude of the frictional force acting on the body?

- Mg
- Mg cos
*θ* - Mg sin
*θ* - Mg tan
*θ*

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**EXERCISE 6: Newton II - Object on a slope**

A force T = 312 N is required to keep a body at rest on a frictionless inclined plane which makes an angle of 35∘ with the horizontal. The forces acting on the body are shown. Calculate the magnitudes of forces P and R, giving your answers to three significant figures.

http://cnx.org/content/m38963/latest/PG11C11_034.png