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7 - Forces and Free Body Diagrams

7 - Forces and Free Body Diagrams. Forces, Energy and Power. Summary Notes page 11. Newton’s 3 laws of motion. How to define the unit of force. How to solve problems using Newton’s Second Law. Newton’s laws of motion.

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7 - Forces and Free Body Diagrams

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  1. 7 - Forces and Free Body Diagrams Forces, Energy and Power Summary Notes page 11.

  2. Newton’s 3 laws of motion. • How to define the unit of force. • How to solve problems using Newton’s Second Law.

  3. Newton’s laws of motion • N1: “An object will remain at rest, or travel at constant speed in a straight line, unless acted upon by an unbalanced force.” • N2: “An object will accelerate when acted upon by an unbalanced force.” • N3: “Every action has an equal and opposite reaction.”

  4. balanced forces ⟶ Constant velocity • Constant velocity ⟶ balanced forces

  5. Definition of the unit of force • N2: “An object will accelerate when acted upon by an unbalanced force.” F = ma • This equation allows us to define the unit of force, the newton, N. • An unbalanced force of 1N is the force required to accelerate a mass of 1kg by 1ms-2. • So: 1 N = 1 kgms-2

  6. Which of the following show balanced forces? a. b. 4N 4N 10N 15N c. d. 10N 16N 8N 14N 10N 4N 4N

  7. Problem Solving Using F=ma • Draw a free-body diagram for the object whose motion is being analysed and indicate all the forces acting on the object. • On the free-body diagram mark the direction of the unbalanced force acting on the object. • Apply F=ma.

  8. Example 1 • Calculate the acceleration of a speed boat of mass 750 kg if the engine produces a forward force of 8000 N and the water exerts a frictional force of 3000 N. F=ma 8000 N 3000 N 5000 =750 × a a = 6.67ms-2 5000 N

  9. Example 2 • Calculate the acceleration of a rocket of mass 3.0 x 106 kg if it experiences an upward force of 4.0 x 107 N. FUB = Fth – mg = ma FUB = 4.0 x 107 - 3.0 x 106 x 9.8 =3.0 x 106 x a a = 3.53ms-2

  10. SAQs page numbers 1 - 7.

  11. SAQ 1. A trolley is pulled with a 16 N force. The force of friction acts against this force and has a magnitude of 4 N. If the mass of the trolley is 1.5 kg, what is the acceleration? SAQ 2. a)A mass of 4 kg has an acceleration of 7.5 ms-2. What is the unbalanced force on the mass ? b) Look at the picture for this question. i. What is the unbalanced force? ii. In which direction is the unbalanced force? iii. What is the acceleration of the mass? c)A force of 240 N pulls a mass of 12 kg. The force of friction is 60 N. What is the unbalanced force and the acceleration?

  12. Example In this problem the mass, acceleration and the force of friction are given. What is the pulling force? Unbalanced force = pulling force - friction ma = pulling force - 5 3 x 6 = pulling force - 5 pulling force = 18 + 5 = 23N to right Note The unbalanced force must be in the same direction as the pulling force because the acceleration is in this direction.

  13. SAQ 3.What is the size of the pulling force? SAQ 4. A pulling force of 36N is acting on a trolley of mass 2kg. The trolley accelerates at 13ms-2. What is the size of the friction force on the trolley ? SAQ 5. A box of 30kg is thrown at 2ms-1 across a factory floor. The force of friction on the box is 45N. What is the deceleration of the box?

  14. SAQ 6. A brick of mass 3kg with a force of 18N acting upon it, initially at rest, accelerates at 3ms-2. Calculate: a)the unbalanced force on the brick; b)the force of friction; c)the bricks speed after 2s. If the 18N force is removed after 2s, calculate: d)the deceleration; e)the time taken for the brick to come to rest.

  15. SAQ 7. A digger is exerting an upward force of 8000 N on a load of mass 600 kg in its bucket. Find the acceleration of the load. (Consider all forces acting.) Structured Problems page 22 Qs 42-49.

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