Applying Newton’s Laws of Motion
 Newton's laws of motion are strong tools to understand the motion of objects with forces acting upon them
 Below are two worked examples demonstrating different situations involving the application of Newton's laws
Worked Example
Two blocks of mass 1 kg and 4 kg respectively are attached by a tight massless rope between them. The 1 kg block sits on the left and the 4 kg block sits on the right. The 1 kg mass has a 100 Newton force applied to it pulling it to the left. What is the acceleration of both blocks and the tension in the rope as they move across a frictionless surface?
A The acceleration is 15 m s–2 to the left and the tension is 20 N
B The acceleration is 20 m s–2 to the left and the tension is 40 N
C The acceleration is 15 m s–2 to the left and the tension is 60 N
D The acceleration is 20 m s–2 to the left and the tension is 80 N
Step 1: Consider the whole of the system

 Together the 1 kg and 4 kg blocks are both being pulled along by the 100 N force (since the rope is tight)
 This is a frictionless flat surface, therefore the only forces in the system are the pulling force and the tension force(s) in the rope
 Therefore the acceleration can be found using Newton's second law
Step 2: Find the acceleration

 Using Newton's second law:
F = m × a

 Rearrange for a
a = F ÷ m
a = 100 ÷ 5 = 20 m s–2 to the left
Step 3: Examine the 4 kg mass only

 Since the system is moving with an acceleration of 20 m s–2 to the left, the force on only the 4 kg mass can be found
F = m × a
F = 4 × 20 = 80 N to the left

 The only force acting on the 4 kg mass is the rope and therefore tension force
 This means there is 20 N pulling force on the 1 kg block only
Step 4: State your answer

 Since acceleration, a = 20 m s–2 to the left and the tension force = 80 N to the left
 Therefore, the answer is option D
Worked Example
A stationary object is subject to a 300 N force towards the left and at 55 degrees leftwards with respect to the vertical and a 450 N force to the right and 35° downwards with respect to the horizontal.Calculate what is the magnitude and direction of the third force that would make this object remain stationary (to the nearest N).
Step 1: Recall Newton's first law

 Newton's First Law states:
A body will remain at rest or move with constant velocity unless acted on by a resultant force

 Therefore, for this object to remain stationary, the resultant force must have a magnitude of 0 Newtons
Step 2: Resolve the 300 N force into its horizontal and vertical components

 The horizontal component can be resolved from:
sin(55°) × 300 = 246 N

 This is directed to the left

 The vertical component can be resolved from:
cos(55°) × 300 = 172 N

 This is in an upwards direction
Step 3: Resolve the 450 N force into its horizontal and vertical components

 The horizontal component can be resolved from:
cos(35°) × 450 = 369 N

 This is directed to the right

 The vertical component can be resolved from:
sin(35°) × 450 = 258 N

 This is in a downwards direction
Step 4: Combine the horizontal components

 The two forces provide 369 N to the right and 246 N to the left
 Therefore, since these are opposing directions:
369 – 246 = 123 N to the right
Step 5: Combine the vertical components

 The two forces provide 258 N downwards and 172 N upwards
 Therefore, since these are opposing directions:
258 – 172 = 86 N downwards
Step 6: Make a right angle triangle using these two force vectors

 There should be a longer size 123 N magnitude vector arrow to the right and then a 86 N magnitude vector arrow downwards
 These can be connected from start to finish by a third vector which is the hypotenuse of this rightangled triangle
Step 7: Use the two vectors magnitudes to find the angle from the horizon

 Since the vectors are to the right and downwards in this rightangle triangle, neither is the hypotenuse
 Therefore the angle from the horizontal downwards can be found by using tan:
tan(θ°) = Opp. ÷ Adj. = 86 ÷ 123
θ° = tan–1(86 ÷ 123) = tan–1(0.699)
θ° ≈ 35°
Step 8: Use Pythagoras theorem or trigonometry to find the magnitude of the resultant force

 Method 1: Using Pythagoras theorem
 c^{2} = b^{2} + a^{2} where b and a are the vector magnitudes for the horizontal and vertical components found so far and c is the hypotenuse magnitude
 Method 1: Using Pythagoras theorem
c^{2} = 86^{2} + 123^{2} = 7396 + 15129 = 22525
c = √22 525 ≈ 150 N

 Method 2: Using trigonometry
 Using the horizontal component and the angle found in step 7, cos can be used
 Method 2: Using trigonometry
cos(35°) = Adj. ÷ Hyp. = 123 ÷ Hyp.


 Therefore:

123 ÷ cos(35°) = Hyp. ≈ 150 N
Step 9: State the final answer

 The third force which would cause this object to remain stationary is 150 N right and 35° downwards from the horizontal
Exam Tip
You are expected to know Newton's three laws of motion from memory and how they apply to physical situations. So be sure to practice and use them without having to review them before carrying out a problem.
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