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Mechanics Chapter 3 Vertical Motion.

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Presentation on theme: "Mechanics Chapter 3 Vertical Motion."β€” Presentation transcript:

1 Mechanics Chapter 3 Vertical Motion

2 3.1. Acceleration due to Gravity
Newton and Galileo both stated: All objects, when dropped, falls towards the Earth in a vertical line with the same constant acceleration, provided there is no air resistance. This acceleration is due to gravity, so gravity itself is a type of acceleration. Gravity is denoted as g. A.K.A. Acceleration of free fall Value of g: 𝑔= 10π‘šπ‘  βˆ’2

3 3.2 Weight When an object of any mass m, is falling with acceleration g, there is a force acting on it. This force is caused by the gravitation pull from the Earth, called the Force of Gravity. This force is also known as Weight. The weight of an object on or near the surface of the Earth is the force of gravity with which the Earth attract it. A.K.A The force of Attraction Weight: π‘Š=π‘šπ‘”=10π‘š

4 Example Determine the Weight of A table of mass 42 kg
A car of mass 1 ton (1 ton = 1000 kg) A sack of rocks with mass 15 lb. (1 kg = 2.2 lb) 42kg X10 = 420N 1000 kg X 10 = N or 10kN 6.8 kg X 10 = 68 N

5 Example Natalie was injured while at sea, and is being winched up to a rescue helicopter. Her mass is 55 kg. Find the tension in the cable when Natalie in being raised up At a steady speed of 4π‘šπ‘  βˆ’1 With an acceleration of 0.8π‘šπ‘  βˆ’2 She is moving at a steady speed, πΉβˆ’π‘…=0, π‘ π‘œ π‘‡βˆ’550=0, β„Žπ‘’π‘›π‘’π‘π‘’ 𝑇=550𝑁 b) There is acceleration, πΉβˆ’π‘…=π‘šπ‘Ž, π‘ π‘œ π‘‡βˆ’550= , β„Žπ‘’π‘›π‘π‘’ 𝑇=594

6 Example A pulley system is used to lift a heavy crate. There are six vertical sections of rope, each having a tension T, and the crate has an upward acceleration a. Find the mass of the crate, expressing your answer in terms of T, a, g. There are several forces, and an object moving with acceleration. Net Force 𝐹 𝑛𝑒𝑑 =6π‘‡βˆ’π‘Š So, 6π‘‡βˆ’π‘šπ‘”=π‘šπ‘Ž Solve for m in terms of T, a, g. So, π‘š= 6𝑇 (𝑔+π‘Ž)

7 Example Machinery of total mass 280 kg is being lowered to the bottom of a mine by means of two ropes attached to a cage of mass 20 kg. For the first 3 seconds of the descent, the tension in each rope is 900 N. Then for a further 16 seconds, the tension in each rope is 1500 N. For the final 8 seconds, the tension in each rope is 1725 N. Find the depth of the mine.

8 3.3 Normal Contact Force When an object is in contact with a surface, there is a force on the object at right angles to the region of contact. This is called the normal contact force. Sometimes called the Normal Reaction. When an object is resting on a horizontal surface the weight and normal force are in equilibrium R=mg

9 Example 1 A book of mass 0.5 kg is placed flat on a horizontal shelf. Find the magnitude of the normal contact force. Weight of the book is 0.5 X 10 = 5 N. The only other force on the book is the normal force, and it also is 5 N.

10 Example 2 A container sits on the dockside waiting to be loaded on to a container ship. The mass of the container is 6000 kg. A cable from a crane is attached to the container. At first, the cable is slack; the tension is then gradually increased until the container rises off the ground. Draw a graph to show the relationship between the normal contact force and the tension in the cable. Three Forces: Weight, Normal, Tension In equilibrium so, R + T – W = 0 When slack, T = 0 and R = 60000, then at point of being lifted R = 0 and T= 60000

11 3.4 Mass and Weight Know the difference between mass and weight.
Mass – the measure of the amount of matter in an object Weight – when an acceleration, like gravity, is applied to mass this creates a force, called Weight.

12 Example 1 Toni, who is a mechanics student, lives on the tenth floor of an apartment building. She has just bough new bathroom scales and decides to try them out by standing on them as she goes up the lift. Initially the scale reads 50 kg. After the doors have closed the reading briefly goes up to 60 kg, but then returns to 50 kg. As the lift nears the tenth floor the reading drops to 35 kg. Find the acceleration of the lift in both stages.

13 Example 2 A heavy mass, m, is suspended from the roof of a lift by a wire. The wire is cut and a spring balance is placed between the two free ends. When the lift is accelerating upwards the reading on the balance is y kg. Find the acceleration and the value for y. A spring balance function is to measure the mass of an object when the upper end is held stainonary. So when the balance reads y kg, the tension in the wire is y N.


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