Neoton’s law

Algorithm for solving problems by Newton’s laws

1. Read the problem

2. Decide which physics equations you should use.

3. Use a rough sketch to help you see how different parts of the problem are related.

4. If the problem is about acceleration then you should also see if it involves polleys

i. Use Fnet = 0 when there is equilibrium

ii. Use Fnet = ma when there is no equilibrium

iii. Use Torquenet = Iα if it involves pulleys

5. If the problem has friction then use static coefficient in the case of friction, and kinetic friction in the case dynamic case.

6. Identify the subject of the forces; what are the forces acting upon?!

7. Draw a coordinate system.

8. Draw all the forces with the proper directions.

9. Make sure all the angles are properly indicated and their measures are referenced correctly to an axis.

10. Write down Newton’s second law of motion as applied to the problem.

11. Now solve the: Depending on the problem, solving Newton’s 2nd law can give two equations two unknowns to solve. As when we solve the x-component and the y-component.

Algorithm for Newtons laws

1

. The type of motion can be identified by the following keywords. General law of motion is s(t ) = at

2

2
+ v0t + s0

• dropped/free fall: means acceleration is a = g = 9.81m/s2
• uniform acceleration/constant acceleration: a = const
• uniform motion/constant speed: a = 0 thus v = const
• initial speed: v0 is given

• initial position: s0 is given

• uniform circular motion: should be given some radius r or some angular velocity ω

• angle of plane or force: means we are to introduce some natural coordinate system to work in

2. How to see the forces.

• all forces acting on body are represented by vectors at the center of mass of body with appropriate direction and
length

• gravitational force is always acting on body, when the system is close to ground; direction of gravitational force is
always by the normal of the ground

• free-body diagrams of bodies in typical situations, where W is the gravitational force

Free Fall Vertical Tension Horizontal Inclined Plane

• in the diagrams by Fdr ag is the dragging force caused by air resistance on body in free fall. If otherwise said it is to
presume zero.

• in diagrams by F f is labelled the friction force which otherwise said should be presumed zero.

• we introduce a coordinate system in which to decompose our vectors.

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. How to decompose vectors.

• vectors have components in respect of given coordinate system

• to add or subtract vectors we add/substract there components

• other way to do it is by triangle law of addition of vectors or parallelogram law

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. How to use Newtons laws

• second Newtons law says that the sum of forces acting on body is equal to the product of its mass and acceleration

• in vector notations this means
∑

i ~Fi = ~F = m~a
• this is shorthand to write that

∑
i (Fi ) j = F j = ma j where with j means the given component

5. How to do the algebra

• we replace with the given information to obtain the unknown

1

1. Write here what the key for the problem and what they do mean.

2. Draw here a free-body diagram.

3. Write here what is given.

4. Write here what is unknown.

5. Introduce coordinate system at 2 and decompose the vectors here.

6. Apply Newtons second law by components here.

7. Work out what is to be found here.

2

Problem We have a body with mass m put on plane with inclined angle θ and zero friction which starts to move from rest.
What is the acceleration of the body.

1. Write here what the key for the problem and what they do mean.

We have a free-body diagram like Inclined Plane where we have the angle and the mass of the body.

2. Draw here a free-body diagram.
3. Write here what is given.

We have the mass m and angle θ and initial velocity v0 = 0(since it starts from rest)
4. Write here what is unknown.

We want to find the acceleration a

5. Introduce coordinate system at 2 and decompose the vectors here.

We introduce coordinate system, such that one of the coordinate axes to be parallel to the hypotenuse of the inclined
plane. The gravitational force ~W is perpendicular to the ground. In this case it hase to components – one normal to
the hypotenuse and one parallel to it. The parallel part is causing the movement of the body. It is easy to show that the
angle between gravitational force and the normal(to the hypotenuse) component is θ too. This means that in this case
the gravitational force has components W1 = mg cosθ; W2 = mg sinθ

6. Apply Newtons second law by components here.

By Newtons second law we have:
∑

i ~Fi = ~F = m~a. In this case, in this coordinate system, we have
onOx ′ : −W1 = mg sinθ = ma ⇒ a = g sinθ
onO y ′ : −W2 +N = 0 ⇒ N = mg cosθ

7. Work out what is to be found here.

From above we see that the acceleration depends on the angle a = g sinθ and in case of vertical inclined plane θ = π/2
we have free fall.

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Problem A child is pulling a sled across a stretch of level ice. The ice has a coefficient of kinetic friction, µk = 0.28. The rope
makes an angle θ = 23◦ with the horizontal, the sled has a mass of 16kg . If the sled is accelerating at a rate of 0.018m/s2, what
is the magnitude of the force applied by the child?

1. Write here what the key for the problem and what they do mean.

We have a child pulling a sled with some force. We know the acceleration a = 0.016m/s2 and angle on which the child
is pulling the slide θ = 23◦ and coefficinet of friction µ= 0.28N /kg

2. Draw here a free-body diagram.
3. Write here what is given.

a = 0.016m/s2, θ = 23◦, µ= 0.28N /kg
4. Write here what is unknown.

The pull force of the child T

5. Introduce coordinate system at 2 and decompose the vectors here.

We are going to use the coordinate system with one axe parallel to the ground and other perpendicular to it. In this case
the pull force T should be decompose on Tx = T cosθ and Ty = T sinθ.

6. Apply Newtons second law by components here.
By Newtons second law we have:
∑

i ~Fi = ~F = m~a. In this case, in this coordinate system, we have
onOx : Tx −F f r = ma ⇒ T cosθ−µk N = ma
onO y : Ty +N −W = 0 ⇒ T sinθ+N −mg = 0

7. Work out what is to be found here.

From the first equation we can express the normal reaction force N = mg −T sinθ. This leads to

T cosθ−µk (mg −T sinθ) = ma

T cosθ+µk T sinθ = ma +µk mg
T (cosθ+µk sinθ) = m(a +µk g )

T = m(a +µk g )
cosθ+µk sinθ

T = 16(0.018+0.28(9.8))
cos23◦+0.28sin23◦ = 43N

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